The dynamics of particles (massive or massless, charged or neutral) around a black hole is among the most important and interesting problems of the black hole astrophysics. These studies not only help us to understand the geometrical structure of spacetimes but also shed light on the high energy phenomenon occurring near the black hole such as formation of jets (which involve escaping particles) and accretion disks (particles orbiting in circular orbits). Due to the presence of strong gravitational and electromagnetic fields, charged particles in general do not follow stable orbits and inter-particle collisions are most common. The aftermath of these collisions among numerous particles lead to various interesting astrophysical phenomenon. There are numerous pieces of astrophysical evidence that a magnetic field might be present in the nearby surrounding of the black holes [1, 2] which support the large scale jets. These jets are most likely the source of cosmic rays and high energy particles coming from nearby galaxies. The origin of this magnetic field is probably the existence of a plasma in the vicinity of a black hole in the form of an accretion disk or a charged gas cloud [3, 4]. The relativistic motion of particles in the conducting matter in the accretion disk can generate the regular magnetic field inside the disk. Therefore near the event horizon of a black hole, it is expected that there exists a very strong magnetic field. To an approximation, it is presumed that this field does not affect the geometry of the black hole but it does affect the motion of the charged particles moving around the black hole [5, 6].

More importantly, a rotating black hole may provide sufficient energy to the particle moving around it due to which the particle may escape to spatial infinity. This physical effect appears to play a crucial role in the ejection of high energy particles from accretion disks around black holes. In the process of ejection of high energy particles, besides the rotation of the black hole, the magnetic field plays an important role [7, 8]. Note that if the black hole is carrying electric charge producing a static electric field (also called Coulomb field), then the mere rotation of the black hole itself induces the magnetic field. Acceleration of the particle by the black hole is generally explained in [9]. Other interesting processes around black holes may include evaporation and phantom energy accretion onto black holes [10–13].

During the motion of a charged particle around a magnetized black hole, it remains under the influence of both gravitational and electromagnetic forces which makes the situation complicated [14, 15]. In the present article, it is considered that a charged particle is orbiting in the innermost stable circular orbit (ISCO) of a slowly rotating Kerr black hole and is suddenly hit by a radially incoming neutral particle. The aftermath of a collision will depend on the energy of the incoming particle which may result in one of the three possible outcomes: the charged particle may escape to infinity; be captured by the black hole, or keep orbiting in ISCO. However, predicting the nature of outcome is compounded by the facts that the particle is charged and interacts with the magnetic field and is frame dragged by the Kerr black hole. It should be noted that the present work is altogether different from the BSW mechanism where two particles (with non-zero angular momentum and high energies) arrive from spatial infinity and collide near the event horizon to generate surplus energy in the center of mass frame [16]. In the literature, the motion of charged particles in ISCO around various black holes has been studied ([17–22] and see references therein).

Here we consider a slowly rotating Kerr black hole which is surrounded by an axially symmetric magnetic field homogeneous at infinity. An almost similar problem was studied for weakly charged rotating black holes in [23]. The main conclusion is that, if the magnetic field is present, then the ISCO is located closer to the black hole horizon. In general, the effect of the black hole rotation on the motion of a neutral particle is the same as the effect of the magnetic field on the motion of a charged particle [24, 25].

To study the escape velocity of a particle from the vicinity of the black hole, in this paper we first consider a neutral particle moving around a slowly rotating Kerr black hole in the absence of a magnetic field and colliding with another particle. For simplicity we consider the motion in the equatorial plane only. Then we consider the same problem for a charged particle in the presence of a magnetic field. We focus on the question under what circumstances the particle can escape from the strong gravitational field to infinity. The magnetic field is homogeneous far from the black hole and the gravitational field is ignorable. Thus, far from the black hole the charged particle moves in a homogeneous magnetic field. If the magnetic field is absent then the equations of motion are fairly simple and can be solved analytically. When a particle moving in a non-uniform magnetic field in the absence of the black hole its motion is chaotic [26, 27]. We are extending previous work [24] for the slowly rotating Kerr black hole.

The outline of the paper is as follows: In Sect. 2 we explain our model and derive an expression for the escape velocity of the neutral particle. In Sect. 3 we derive the equations of motion of the charged particle moving around a slowly rotating weakly magnetized Kerr black hole. In Sect. 4 we give the dimensionless form of the equations. Trajectories for escape energy and escape velocity of the particle are discussed in Sects. 5 and 6, respectively. Summary and conclusion are presented in Sect. 7. Throughout we use the sign convention (+,-,-,-)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(+,-,-,-)$$\end{document} and units where c=1,G=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$c=1,G=1$$\end{document}.

We start with the simple case of calculating the escape velocity when the particle is neutral and the magnetic field is absent. The Kerr metric is given by [28]1ds2=Δ-a2sin2θρ2dt2+4Marsin2θρ2dϕdt-ρ2Δdr2-ρ2dθ2-Asin2θρ2dϕ2,Δ≡r2-2Mr+a2,ρ2≡r2+a2cos2θ,A≡(r2+a2)2-a2Δsin2θ,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned}&\mathrm{d}s^{2}=\frac{\Delta -a^{2}\sin ^{2}\theta }{\rho ^{2}}\mathrm{d}t^{2}+\frac{4Mar\sin ^{2}\theta }{\rho ^{2}}\mathrm{d}\phi \mathrm{d}t\nonumber \\&\qquad \qquad -\,\frac{\rho ^{2}}{\Delta }\mathrm{d}r^{2} -\rho ^{2}\mathrm{d}\theta ^{2}-\frac{A\sin ^{2}\theta }{\rho ^{2}}\mathrm{d}\phi ^{2}, \nonumber \\&\Delta \equiv r^{2}-2Mr+a^{2}, \ \ \rho ^{2}\equiv r^{2}+a^{2}\cos ^{2}\theta , \nonumber \\&A \equiv (r^{2}+a^{2})^{2}-a^{2}\Delta \sin ^{2}\theta , \end{aligned}$$\end{document}where M\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M$$\end{document} is the mass and a\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a$$\end{document} is the spin of the black hole and interpreted as the angular momentum per unit mass of the black hole a=LM\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a=\frac{L}{M}$$\end{document}. The horizons of the Kerr metric are obtained by solving2Δ(r)=r2+a2-2Mr=0.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Delta (r)=r^{2}+a^{2}-2Mr=0. \end{aligned}$$\end{document}From the above equation we get two values of r\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r$$\end{document}:3r+=M+M2-a2,r-=M-M2-a2.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} r_{+}=M+\sqrt{M^{2}-a^{2}}, \quad r_{-}=M-\sqrt{M^{2}-a^{2}}. \end{aligned}$$\end{document}Note that Δ>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Delta >0$$\end{document} for r>r+\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r>r_{+}$$\end{document} and r<r-\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r<r_{-}$$\end{document} and Δ<0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Delta <0$$\end{document} for r-<r<r+\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r_{-}<r<r_{+}$$\end{document} [29]. The region r=r+\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r=r_{+}$$\end{document} represents the event horizon while r-\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r_-$$\end{document} is termed the Cauchy horizon. Further r=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r=0$$\end{document} and θ=π2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\theta =\frac{\pi }{2}$$\end{document} is the location of a curvature ring-like singularity in the Kerr spacetime.

In the literature, slowly rotating Kerr black holes have been investigated for numerous astrophysical processes including retro-MACHOS [30], particle acceleration via the BSW mechanism [31], thin accretion disk and accretion rates [32, 33], to list a few. Hence we consider the slowly rotating black hole and neglect the terms involving a2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a^{2}$$\end{document}. The line element in (1) becomes4ds2=1-rgrdt2+4aMsin2θrdϕdt-11-rgrdr2-r2dθ2-r2sin2θdϕ2.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathrm{d}s^2&= \left( 1-\frac{r_{\mathrm{g}}}{r}\right) \mathrm{d}t^2+\frac{4aM\sin ^{2}\theta }{r}\mathrm{d}\phi \mathrm{d}t\nonumber \\&-\frac{1}{1-\frac{r_{\mathrm{g}}}{r}}\mathrm{d}r^2-r^2\mathrm{d}\theta ^2-r^2\sin ^2\theta \mathrm{d}\phi ^2. \end{aligned}$$\end{document}Here rg=2M\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r_{\mathrm{g}}=2M$$\end{document} is the gravitational radius of the slowly rotating Kerr black hole just like the Schwarzschild black hole (note that for a slowly rotating Kerr and Schwarzschild black hole the horizon occurs at r=rg\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r=r_{\mathrm{g}}$$\end{document}). Clearly the metric (4) is stationary but non-static since dt→-dt\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm{d}t\rightarrow -\mathrm{d}t$$\end{document}, changes the signature of the metric. The metric is also axially symmetric (invariance under dθ→-dθ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm{d}\theta \rightarrow -\mathrm{d}\theta $$\end{document}).

In terms of Lagrangian mechanics (L=gμνx˙μx˙ν\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {L}=g_{\mu \nu }\dot{x}^\mu \dot{x}^\nu $$\end{document}), the t\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$t$$\end{document} and ϕ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\phi $$\end{document} coordinates are cyclic, which leads to two conserved quantities, namely energy and angular momentum, with the corresponding Noether symmetry generators5ξ(t)=ξ(t)μ∂μ=∂∂t,ξ(ϕ)=ξ(ϕ)μ∂μ=∂∂ϕ.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \xi _{(t)}= \xi _{(t)}^\mu \partial _\mu =\frac{\partial }{\partial t}, \qquad \xi _{(\phi )}=\xi _{(\phi )}^\mu \partial _\mu =\frac{\partial }{\partial \phi }. \end{aligned}$$\end{document}This shows that the black hole metric is invariant under time translation and rotation around the symmetry axis. The corresponding conserved quantities are the energy E\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}$$\end{document} per unit mass and azimuthal angular momentum Lz\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L_{z}$$\end{document} per unit mass16t˙=r3E+aLzrgr2(r-rg),ϕ˙=1r2argE(r-rg)+Lzsin2θ.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \dot{t}&=\frac{r^{3}\mathcal {E}+aL_{z}r_{\mathrm{g}}}{r^{2}(r-r_{\mathrm{g}})},\nonumber \\ \dot{\phi }&=\frac{1}{r^{2}}\left( \frac{ar_{\mathrm{g}}\mathcal {E}}{(r-r_{\mathrm{g}})}+\frac{L_{z}}{\sin ^{2}\theta }\right) . \end{aligned}$$\end{document}From the astrophysical perspective, it is well known that particles orbit a rotating black hole in the equatorial plane [34]. Therefore we choose θ=π2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\theta =\frac{\pi }{2}$$\end{document} to get7t˙=r3E+aLzrgr2(r-rg),ϕ˙=1r2argE(r-rg)+Lz.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \dot{t}&=\frac{r^{3}\mathcal {E}+aL_{z}r_{\mathrm{g}}}{r^{2}(r-r_{\mathrm{g}})},\nonumber \\ \dot{\phi }&=\frac{1}{r^{2}}\left( \frac{ar_{\mathrm{g}}\mathcal {E}}{(r-r_{\mathrm{g}})}+L_{z}\right) . \end{aligned}$$\end{document}Throughout this paper the over dot represents differentiation with respect to proper time τ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\tau $$\end{document}.

Using the normalization condition, uμuμ=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$u^{\mu }u_{\mu }=1$$\end{document}, we get the equation of motion8r˙2=(Er2-aLz)2r4-r2-rgrr4(r2+Lz2-2aELz).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \dot{r}^{2}=\frac{(\mathcal {E} r^{2}- aL_{z})^2}{r^{4}}-\frac{r^{2}-r_{\mathrm{g}}r}{r^{4}}(r^{2}+L_{z}^{2}-2a\mathcal {E} L_{z}). \end{aligned}$$\end{document}At the turning points r˙=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\dot{r}=0$$\end{document}, Eq. (8) is quadratic in E\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}$$\end{document} whose solution is9E=aLzrg±r5(r-rg)+Lz2(r4-r3rg+a2rg2)r3,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathcal {E}=\frac{aL_{z}r_{\mathrm{g}}\pm \sqrt{r^{5}(r-r_{\mathrm{g}})+L_{z}^{2}(r^{4}-r^{3}r_{\mathrm{g}}+a^{2}r_{\mathrm{g}}^{2})}}{r^{3}}, \end{aligned}$$\end{document}which gives E=Veff\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}=V_\text {eff}$$\end{document}, the effective potential. The condition r˙=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\dot{r}=0$$\end{document} is termed the turning point because it gives the location at which an incoming particle turns around from the neighborhood of the gravitating source [35]. As we are considering only the positive energy we will consider only the positive sign before the square root in Eq. (9) for all the further calculation. Equations (8) and (9) hold for equatorial plane only. It can be seen from (9) that E→1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}\rightarrow 1$$\end{document} for r→∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r\rightarrow \infty $$\end{document}. Therefore the minimum energy for the particle to escape from the vicinity of the black hole is 1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$1$$\end{document}.

Consider a particle in ISCO, where ro\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r_{\mathrm{o}}$$\end{document} is the local minimum (which is also the convolution point) of the effective potential [29]. The corresponding energy and azimuthal angular momentum are given by [29, 36] after neglecting terms involving a2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a^{2}$$\end{document}; we have10Lzo=±rgro±a2rgro2ro-3rg∓2a2rgro,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} L_{z\mathrm{o}}=\pm \frac{\sqrt{r_{\mathrm{g}}}\left( r_{\mathrm{o}}\pm a\sqrt{\frac{2r_{\mathrm{g}}}{r_{\mathrm{o}}}}\right) }{\sqrt{2r_{\mathrm{o}}-3r_{\mathrm{g}}\mp 2a\sqrt{\frac{2r_{\mathrm{g}}}{r_{\mathrm{o}}}}}}, \end{aligned}$$\end{document}11Eo=1-rgr∓arrg2r1-3rg2r∓ar2rgr.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathcal {E}_{\mathrm{o}}=\frac{1-\frac{r_{\mathrm{g}}}{r}\mp \frac{a}{r}\sqrt{\frac{r_{\mathrm{g}}}{2r}}}{\sqrt{1-\frac{3r_{\mathrm{g}}}{2r}}\mp \frac{a}{r}\sqrt{\frac{2r_{\mathrm{g}}}{r}}}. \end{aligned}$$\end{document}Now consider the particle in the ISCO which collides with another incoming particle. After collision between these particles, three cases are possible for the motion of the particle: (i) bound motion, (ii) capture by the black hole, (iii) escape to infinity. The result will depend on the collision process. For a small change in energy and momentum, the orbit of the particle will be slightly perturbed. For a large change in energy and angular momentum, the particle can either be captured by the black hole or escape to infinity.

After the collision the particle should have new values of energy and momentum E\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}$$\end{document}, Lz\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L_{z}$$\end{document}, and the total angular momentum L2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^{2}$$\end{document}. We simplify the problem by applying the following conditions: (i) the azimuthal angular momentum is fixed, (ii) the initial radial velocity remains the same after the collision. Under these conditions only the energy of the particle can determine its motion. After collision the particle acquires an escape velocity (v⊥)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(v_{\bot })$$\end{document} in an orthogonal direction of the equatorial plane [37]. The square of the total angular momentum of the particle after a collision is given by12L2=r4θ2˙+r4sin2θϕ˙2.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} L^{2}=r^{4}\dot{\theta ^{2}}+r^{4}\sin ^{2}\theta \dot{\phi }^{2}. \end{aligned}$$\end{document}Putting the value of ϕ˙\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\dot{\phi }$$\end{document} from Eq. (6) in Eq. (12) we have13L2=r2v⊥2+sin2θargEor-rg+Lzosin2θ2.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} L^{2}=r^{2}v_{\perp }^{2}+\sin ^{2}\theta \left( \frac{ar_{\mathrm{g}}\mathcal {E}_{\mathrm{o}}}{r-r_{\mathrm{g}}}+\frac{L_{z\mathrm{o}}}{\sin ^{2}\theta }\right) ^{2}. \end{aligned}$$\end{document}Here we denote v≡-rθ˙o\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$v\equiv -r\dot{\theta }_{\mathrm{o}}$$\end{document}. Note that L2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^2$$\end{document} is not the integral of motion. It is conserved for a=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a=0$$\end{document}, i.e. in the spherically symmetric case. However, now the metric is axially symmetric, therefore only the Lz\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L_z$$\end{document} component is conserved. In a flat spacetime, all three components Lx\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L_x$$\end{document}, Ly\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L_y$$\end{document}, Lz\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L_z$$\end{document} are conserved, and so is the square of the total angular momentum. The angular momentum Lzo\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L_{z\mathrm{o}}$$\end{document} and energy Eo\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}_{\mathrm{o}}$$\end{document} appearing in (13) are given by (10) and (11), which provide the necessary corrections due to the spin of the black hole.

From Eqs. (9) and (13), the angular momentum and the energy of the particle after the collision become14L2=ro2v⊥2+argEoro-rg+Lzo2,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned}&L^{2}=r_{\mathrm{o}}^{2}v_{\perp }^{2}+\left( \frac{ar_{\mathrm{g}}\mathcal {E}_{\mathrm{o}}}{r_{\mathrm{o}}-r_{\mathrm{g}}}+L_{z\mathrm{o}}\right) ^{2},\end{aligned}$$\end{document}15Enew=aLrg+ro5(ro-rg)+L2(ro4-ro3rg+a2rg2)ro3.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned}&\mathcal {E_\text {new}}=\frac{aLr_{\mathrm{g}}+\sqrt{r_{\mathrm{o}}^{5}(r_{\mathrm{o}}-r_{\mathrm{g}}) +L^{2}(r_{\mathrm{o}}^{4}-r_{\mathrm{o}}^{3}r_{\mathrm{g}}+a^{2}r_{\mathrm{g}}^{2})}}{r_{\mathrm{o}}^{3}}.\nonumber \\ \end{aligned}$$\end{document}These values of the angular momentum and energy are greater than their values before the collision. Physically it means that the energy of the particle exceeds its rest mass energy. We have mentioned above that all the orbits with Enew≥1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}_\text {new}\ge 1$$\end{document} are unbounded in the sense that the particle escapes to infinity. Conversely for Enew<1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}_\text {new}<1$$\end{document}, the particle cannot escape to infinity (the orbits are always bounded).

Therefore the particle escapes to infinity if Enew≥1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}_\text {new}\ge 1$$\end{document}, or16v⊥≥±r(rg-r)(Lz(r-rg)+arg(Eo-1))+r2rg(r-rg)2(r3+rg(a2-r2-2a2Eo))r2(r-rg)2.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} v_{\perp }\ge \pm \frac{r(r_{\mathrm{g}}-r)(L_{z}(r-r_{\mathrm{g}})+ar_{\mathrm{g}}(\mathcal {E}_{\mathrm{o}}-1))+\sqrt{r^{2}r_{\mathrm{g}}(r-r_{\mathrm{g}})^{2}(r^{3} +r_{\mathrm{g}}(a^{2}-r^{2}-2a^{2}\mathcal {E}_{\mathrm{o}}))}}{r^{2}(r-r_{\mathrm{g}})^{2}}. \end{aligned}$$\end{document}The particle escape condition is |v|≥v⊥\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$|v|\ge v_{\perp }$$\end{document}, i.e. the magnitude of the velocity should be greater than any orthogonal velocity.

Here we investigate the motion of a charged particle (electric charge q\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$q$$\end{document}) in the presence of a magnetic field in the exterior of the slowly rotating Kerr black hole. The Killing equation is17□ξμ=0,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \square \xi ^{\mu }=0, \end{aligned}$$\end{document}where ξμ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\xi ^{\mu }$$\end{document} is a Killing vector. Note that (17) follows from the result: a Killing vector in a vacuum spacetime generates a solution of the Maxwell equations i.e. Fμν=-2ξμ;ν\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$F_{\mu \nu }=-2\xi _{\mu ;\nu }$$\end{document}. From F;νμν=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$F^{\mu \nu }_{~~;\nu }=0$$\end{document}, it follows that -2ξ;νμ;ν=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$-2\xi ^{\mu ;\nu }_{~~;\nu }=0$$\end{document}. Thus (17) coincides with the Maxwell equation for the 4-potential Aμ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A^{\mu }$$\end{document} in the Lorenz gauge A;μμ=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A^{\mu }_{\ \ ;\mu }=0$$\end{document}. The special choice for Aμ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A^{\mu }$$\end{document} is [2, 23]18Aμ=aB,0,0,B2,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} A^{\mu }= \left( a\mathcal {B},0,0,\frac{\mathcal {B}}{2} \right) , \end{aligned}$$\end{document}where B\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {B}$$\end{document} is the magnetic field strength. The 4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$4$$\end{document}-potential is invariant under the symmetries which correspond to the Killing vectors, i.e.,19LξAμ=Aμ,νξν+Aνξ,μν=0.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} L_{\xi }A_{\mu }=A_{\mu ,\nu }\xi ^{\nu }+A_{\nu }\xi ^{\nu }_{,\mu }=0. \end{aligned}$$\end{document}A magnetic field vector is defined as20Bμ=-12eμνλσFλσuν,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathcal {B}^{\mu }=-\frac{1}{2}\mathrm{e}^{\mu \nu \lambda \sigma }F_{\lambda \sigma }u_{\nu }, \end{aligned}$$\end{document}where21eμνλσ=ϵμνλσ-g,ϵ0123=1,g=det(gμν).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathrm{e}^{\mu \nu \lambda \sigma }=\frac{\epsilon ^{\mu \nu \lambda \sigma }}{\sqrt{-g}},\ \ \epsilon _{0123}=1,\ \ g=\mathrm{det}(g_{\mu \nu }). \end{aligned}$$\end{document}In (21) ϵμνλσ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\epsilon ^{\mu \nu \lambda \sigma }$$\end{document} is the Levi-Civita symbol and the Maxwell tensor is defined as22Fμν=Aν;μ-Aμ;ν.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} F_{\mu \nu }=A_{\nu ;\mu }-A_{\mu ;\nu }. \end{aligned}$$\end{document}For a local observer at rest we have23uμ=1(1-rgr)+4aM1-rgrr2sinθ,0,0,1rsinθ1+4aMr2sinθ1-rgr.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned}&u^{\mu }=\left( \frac{1}{\sqrt{(1-\frac{r_{\mathrm{g}}}{r})+\frac{4aM\sqrt{\left( 1-\frac{r_{\mathrm{g}}}{r}\right) }}{r^{2}\sin \theta }}},0,0,\nonumber \right. \\&\quad \left. \frac{1}{r\sin \theta \sqrt{\left( 1+\frac{4aM}{r^{2}\sin \theta \sqrt{\left( 1-\frac{r_{\mathrm{g}}}{r}\right) }}\right) }}\right) . \end{aligned}$$\end{document}From (20)–(23) we obtain the components of the magnetic field,24Bμ=B0,cosθ(1-rgr)(1-rgr)+2rga(1-rgr)r2sinθ+rgasinθcosθr(1-rgr)rsinθ(1+2rgar2sinθ(1-rgr)),-sinθ(1-rgr)r(1-rgr)+2rga(1-rgr)r2sinθ,0.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathcal {B}^{\mu }&= \mathcal {B}\left( 0,\cos \theta \left( \frac{\bigg (1-\frac{r_{\mathrm{g}}}{r}\bigg )}{\sqrt{\bigg (1-\frac{r_{\mathrm{g}}}{r}\bigg )+ \frac{2r_{\mathrm{g}}a\sqrt{\bigg (1-\frac{r_{\mathrm{g}}}{r}\bigg )}}{r^{2}\sin \theta }}}\right) \right. \nonumber \\&\left. +\,\,\frac{r_{\mathrm{g}}a\sin \theta \cos \theta }{r} \!\left( \!\frac{\bigg (1-\frac{r_{\mathrm{g}}}{r}\bigg )}{r\sin \theta \!\sqrt{\bigg (\!1\!+\!\frac{2r_{\mathrm{g}}a}{r^{2}\sin \theta \!\sqrt{\bigg (\!1-\frac{r_{\mathrm{g}}}{r}\!\bigg )}}\!\bigg )}} \!\right) \!,\right. \nonumber \\&\left. -\frac{\sin \theta \bigg (1-\frac{r_{\mathrm{g}}}{r}\bigg )}{r\sqrt{\bigg (1-\frac{r_{\mathrm{g}}}{r}\bigg )+\frac{2r_{\mathrm{g}}a\sqrt{\bigg (1-\frac{r_{\mathrm{g}}}{r}\bigg )}}{r^{2}\sin \theta }}},0\right) . \end{aligned}$$\end{document}For the equatorial plane only, the third component of the magnetic field will survive. Hence Eq. (24) becomes25Bμ=B0,0,-1-rgrr1-rgr+2rga1-rgrr2,0.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathcal {B}^{\mu }&= \mathcal {B}\left( 0,0, -\frac{\left( 1-\frac{r_{\mathrm{g}}}{r}\right) }{r\sqrt{\left( 1-\frac{r_{\mathrm{g}}}{r}\right) +\frac{2r_{\mathrm{g}}a\sqrt{\left( 1-\frac{r_{\mathrm{g}}}{r}\right) }}{r^{2}}}},0\right) . \end{aligned}$$\end{document}The Lagrangian of the particle of mass m\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$m$$\end{document} and charge q\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$q$$\end{document} moving in an external magnetic field in a curved spacetime is [38]26L=12gμνx˙μx˙ν+qAμmx˙μ,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathcal {L}=\frac{1}{2}g_{\mu \nu }\dot{x}^{\mu }\dot{x}^{\nu }+\frac{qA_{\mu }}{m}\dot{x}^{\mu }, \end{aligned}$$\end{document}and the generalized 4-momentum of the particle is Pμ=muμ+qAμ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$P_{\mu }=mu_{\mu }+qA_{\mu }$$\end{document}. The constants of motion are27t˙=r3E+aLzrgr2(r-rg)-2aB,ϕ˙=1r2argE(r-rg)+Lzsin2θ-B.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \dot{t}&=\frac{r^{3}\mathcal {E}+aL_{z}r_{\mathrm{g}}}{r^{2}(r-r_{\mathrm{g}})}-2aB, \nonumber \\ \dot{\phi }&=\frac{1}{r^{2}}\left( \frac{ar_{\mathrm{g}}\mathcal {E}}{(r-r_{\mathrm{g}})}+\frac{L_{z}}{\sin ^{2}\theta }\right) -B. \end{aligned}$$\end{document}For the equatorial plane θ=π2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\theta =\frac{\pi }{2}$$\end{document} the above integrals of motion become28t˙=r3E+aLzrgr2(r-rg)-2aB,ϕ˙=1r2argE(r-rg)+Lz-B.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \dot{t}&=\frac{r^{3}\mathcal {E}+aL_{z}r_{\mathrm{g}}}{r^{2}(r-r_{\mathrm{g}})}-2aB, \nonumber \\ \dot{\phi }&=\frac{1}{r^{2}}\left( \frac{ar_{\mathrm{g}}\mathcal {E}}{(r-r_{\mathrm{g}})}+L_{z}\right) -B. \end{aligned}$$\end{document}Here we denote29B≡qB2m.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} B\equiv \frac{q\mathcal {B}}{2m}. \end{aligned}$$\end{document}Using the values of t˙\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\dot{t}$$\end{document} and ϕ˙\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\dot{\phi }$$\end{document} and neglecting the terms involving a2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a^{2}$$\end{document}, Eq. (26) yields30L=12r2(r-rg)2[4Br2Lz(rg-r)+Lz2(rg-r)+r2(Brg(3Br2+2aE)r(E2-3B2r2-r˙2))].\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathcal {L}&= \frac{1}{2r^{2}(r-r_{\mathrm{g}})^{2}}[4Br^{2}L_{z}(r_{\mathrm{g}}-r)+L_{z}^{2}(r_{\mathrm{g}}-r) \nonumber \\&+\,r^{2}(Br_{\mathrm{g}}(3Br^{2}+2a\mathcal {E})r(\mathcal {E}^{2}-3B^{2}r^{2}-\dot{r}^{2}))]. \end{aligned}$$\end{document}By using the above Lagrangian in the Euler–Lagrange equation, which is defined as31ddτ∂L∂x˙-∂L∂x=0,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}\tau }\left( \frac{\partial \mathcal {L}}{\partial \dot{x}}\right) -\frac{\partial \mathcal {L}}{\partial {x}}=0, \end{aligned}$$\end{document}we get32r¨=BaErgr(r-rg)+12r4(r-rg)[6B2r6-2Lz2(r-rg)2+r3rg(-E2+6B2rrg+r˙2-12B2r2)].\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \ddot{r}&= \frac{Ba\mathcal {E}r_{\mathrm{g}}}{r(r-r_{\mathrm{g}})} +\frac{1}{2r^{4}(r-r_{\mathrm{g}})}[6B^{2}r^{6}-2L_{z}^{2}(r-r_{\mathrm{g}})^{2} \nonumber \\&+\,r^{3}r_{\mathrm{g}}(-\mathcal {E}^{2}+6B^{2}rr_{\mathrm{g}}+\dot{r}^{2}-12B^{2}r^{2})]. \end{aligned}$$\end{document}Following the procedure of Sect. 2, using the normalization condition, uμuμ=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$u^{\mu }u_{\mu }=1$$\end{document}, and using the value of the new constants of motion, see (28), we obtain33E=1ro6(ro-rg)[2aBro7+argro3(2Bro2(rg-2ro)+Lz(rg-ro))±(a2ro6(ro-rg)2(rg(Lz+2Bro2)-2Bro3)2+ro9(ro-rg)3(ro2+(Lz+Bro2)2))12].\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathcal {E}&= \frac{1}{r_{\mathrm{o}}^{6}(r_{\mathrm{o}}-r_{\mathrm{g}})}[2aBr_{\mathrm{o}}^{7}+ar_{\mathrm{g}}r_{\mathrm{o}}^{3}(2Br_{\mathrm{o}}^{2}(r_{\mathrm{g}}-2r_{\mathrm{o}})\nonumber \\&+\,L_{z}(r_{\mathrm{g}}-r_{\mathrm{o}})) \nonumber \\&\pm \,(a^{2}r_{\mathrm{o}}^{6}(r_{\mathrm{o}}-r_{\mathrm{g}})^{2}(r_{\mathrm{g}}(L_{z}+2Br_{\mathrm{o}}^{2})-2Br_{\mathrm{o}}^{3})^{2} \nonumber \\&+\,r_{\mathrm{o}}^{9}(r_{\mathrm{o}}-r_{\mathrm{g}})^{3}(r_{\mathrm{o}}^{2}+(L_{z}+Br_{\mathrm{o}}^{2})^{2}))^{\frac{1}{2}}]. \end{aligned}$$\end{document}If (33) is satisfied initially (at the time of collision), then it is always valid (throughout the motion), provided that r(τ)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r(\tau )$$\end{document} is controlled by (32).

The system (26)–(33) is invariant with respect to reflection (θ→π-θ)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(\theta \rightarrow \pi -\theta )$$\end{document}. This transformation retains the initial position of the particle and changes (v⊥→-v⊥)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(v_{\perp }\rightarrow -v_{\perp })$$\end{document} as it is defined by v⊥≡-rθo˙\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$v_{\perp }\equiv -r\dot{\theta _{\mathrm{o}}}$$\end{document}. Therefore, it is sufficient to consider only the positive value of (v⊥)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(v_{\perp })$$\end{document}.

To perform the numerical analysis, it is convenient to convert Eqs. (32) and (33) to dimensionless form by introducing the following dimensionless quantities:34σ=τrg,ρ=rrg,ℓ=Lzrg,b=Brg.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \sigma =\frac{\tau }{r_{\mathrm{g}}},\ \rho =\frac{r}{r_{\mathrm{g}}},\ \ell =\frac{L_{z}}{r_{\mathrm{g}}},\ b=Br_{\mathrm{g}}. \end{aligned}$$\end{document}Equation (33) now becomes35Eo=1ρo6(ρo-1)[aρo3(1-ρo)(ℓ-2bρo2(ρo-1))+(ρo6(ρo-1)2(a2(ℓ-2bρo2(ρo-1))2)+ρo3(ρo-1)(ρo2+(ℓ+bρo2)2))12].\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathcal {E}_{\mathrm{o}}&= \frac{1}{\rho _{\mathrm{o}}^{6}(\rho _{\mathrm{o}}-1)}[a\rho _{\mathrm{o}}^{3}(1-\rho _{\mathrm{o}})(\ell -2b\rho _{\mathrm{o}}^{2}(\rho _{\mathrm{o}}-1)) \nonumber \\&+\,(\rho _{\mathrm{o}}^{6}(\rho _{\mathrm{o}}-1)^{2}(a^{2}(\ell -2b\rho _{\mathrm{o}}^{2}(\rho _{\mathrm{o}}-1))^{2}) \nonumber \\&+\,\rho _{\mathrm{o}}^{3}(\rho _{\mathrm{o}}-1)(\rho _{\mathrm{o}}^{2}+(\ell +b\rho _{\mathrm{o}}^{2})^{2}) )^{\frac{1}{2}}]. \end{aligned}$$\end{document}The magnetic field is zero at ρ→∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho \rightarrow \infty $$\end{document}. Therefore from Eq. (35) as ρ→∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho \rightarrow \infty $$\end{document}, we have E→1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}\rightarrow 1$$\end{document}.

The dimensionless form of Eq. (32) is36d2ρdσ2=12ρ4(ρ-1)[ρ3(2aEb+6E2b2ρ(ρ-1)2)-2ℓ(ρ-1)2+ρ3dρdσ].\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \frac{\mathrm{d}^{2}\rho }{\mathrm{d}\sigma ^{2}}&= \frac{1}{2\rho ^{4}(\rho -1)}\bigg [\rho ^{3}(2a\mathcal {E}b+6\mathcal {E}^{2}b^{2}\rho (\rho -1)^{2})\nonumber \\&-\,2\ell (\rho -1)^{2}+\rho ^{3}\frac{\mathrm{d}\rho }{\mathrm{d}\sigma }\bigg ]. \end{aligned}$$\end{document}We solved Eq. (36) numerically by using the built in command NDSolve of Mathematica. As ISCO exists at r=3rg\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r=3r_{\mathrm{g}}$$\end{document}, and using ρ=rrg\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho =\frac{r}{r_{\mathrm{g}}}$$\end{document} and σ=τrg\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sigma =\frac{\tau }{r_{\mathrm{g}}}$$\end{document}, our initial conditions for solving (36) become ρ(1)=3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho (1)=3$$\end{document} and ρ˙(1)=3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\dot{\rho }(1)=3$$\end{document}. We get the interpolating function ρ(σ)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho (\sigma )$$\end{document} as the solution of the Eq. (36) which we plotted in Fig. 1 against σ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sigma $$\end{document}. In Fig. 2 we have plotted the radial velocity (derivative of the interpolating function) vs. σ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sigma $$\end{document}, which shows that the particle will escape to infinity according to the initial conditions.Fig. 1

The graph for ρ(σ)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho (\sigma )$$\end{document} vs. σ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sigma $$\end{document}. Here E=1,q=1,b=0.5,ℓ=2,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}=1, q=1, b=0.5, \ell =2,$$\end{document} and a=0.1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a=0.1$$\end{document}

Here we investigate the dynamics of the particle for the positive energy E+\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}_{+}$$\end{document}. Particles with negative energy exist only inside the static limit surface (rst=2m)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(r_{st}=2m)$$\end{document} orbiting in the retrograde orbits and do not have the chance to escape. The equation for the rotational (angular) variable ϕ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\phi $$\end{document} is40dϕdσ=ℓρ2-b+aEρ3(1-ρ).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \frac{\mathrm{d}\phi }{\mathrm{d}\sigma }=\frac{\ell }{\rho ^{2}}-b+\frac{a\mathcal {E}}{\rho ^{3}(1-\rho )}. \end{aligned}$$\end{document}The Lorentz force acting on the massive charged particle is attractive when dϕ/dσ<0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm{d}\phi /\mathrm{d}\sigma <0$$\end{document} and vice versa. All of Figs. (3, 4, 5, 6, 7, 8) correspond to Eq. (35). In Fig. 3, the shaded region corresponds to an unbound motion, while the unshaded region refers to bounded trajectories of the particle. The curved line represents the minimum energy required for the particle to escape from the vicinity of the black hole. It can be seen from Fig. 4 that for large values of the angular momentum, the plot is similar to the effective potential of a Schwarzschild black hole [24]. In Fig. 4, Emax\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}_\text {max}$$\end{document} corresponds to an unstable circular orbit and Emin\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}_\text {min}$$\end{document} refers to ISCO. Fig. 3

The effective potential E\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}$$\end{document} as a function of ρ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho $$\end{document} for ℓ=5\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\ell =5$$\end{document}, b=0.5\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$b=0.5$$\end{document}, and a=0.1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a=0.1$$\end{document}

For all the figures of the escape velocity we have denoted v⊥≡vesc\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$v_{\perp }\equiv v_\text {esc}$$\end{document}. From Eq. (38) we calculate the escape velocity by substituting E=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {E}=1$$\end{document}. Figures 9, 10, 11, and 12 correspond to Eq. (39). In Fig. 9, the shaded region corresponds to the escape velocity of the particle and the solid curve represents the minimum velocity required to escape from the vicinity of the black hole to infinity. The unshaded region represents the bound motion around the black hole. In Fig. 10 the shaded region corresponds to the escape velocity of the particle and the solid curve represents the minimum velocity required to escape from the vicinity of the black hole. The unshaded region represents the bound motion around the black hole. Fig. 9

The escape velocity against ρ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho $$\end{document} for ℓ=5\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\ell =5$$\end{document}, b=0.5\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$b=0.5$$\end{document} and a=0.1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a=0.1$$\end{document}

We have studied the dynamics of a neutral and a charged particle around the slowly rotating Kerr black hole which is immersed in a magnetic field. Therefore the particle is under the influence of both gravitational and electromagnetic forces. We have obtained the equations of motion by using Lagrangian formalism. We have derived the expression for the magnetic field present in the vicinity of a slowly rotating Kerr black hole. We have calculated the minimum energy for a particle to escape from ISCO to infinity. With zero spin i.e. a=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a=0$$\end{document}, our results reduce to the case of the Schwarzschild black hole [15].

The behavior of the effective potential and escape velocity against magnetic field and angular momentum are discussed in detail. It is shown in Figs. 4, 9, and 10 under what conditions particle can escape from the vicinity of the black hole to spatial infinity. For larger values of the angular momentum, the behavior of the effective potential is similar to that of the Schwarzschild black hole [15]. It is concluded that the magnetic field largely affects the motion of the particle in the vicinity of the black hole. This effect decreases far away from the black hole. It is found that as the value of the magnetic field parameter is increased, the local minimum of the effective potential shifted towards the horizon, as shown in Fig. 8. This indicates that the ISCO shrinks as the strength of the magnetic field increases. It is concluded from Figs. 5 and 12 that if the particle has a large value angular momentum ℓ+\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\ell _{+}$$\end{document}, then it can escape easily as compared to a particle with smaller angular momentum ℓ+\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\ell _{+}$$\end{document}. Figure 7 shows that for an attractive Lorentz force (ℓ-)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(\ell _{-})$$\end{document} the stability is larger in comparison with a repulsive Lorentz force (ℓ+)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(\ell _{+})$$\end{document}.

The escape velocity vesc\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$v_\mathrm{esc}$$\end{document}, for different values of the magnetic field b\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$b$$\end{document} is plotted in Fig. 11. It is found that due to the presence of the magnetic field in the vicinity of the black hole the escape velocity of the particle increases. Therefore we found that the possibility of the particle to escape from the vicinity of the black hole to infinity is greater in the presence of a magnetic field as compared to the case when the magnetic field is absent, b=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$b=0$$\end{document}.