JEE Challenger
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PhyMedAdv 2007 • Paper 2

Trajectory of Charged Particle in Discontinuous Magnetic Field

A magnetic field B⃗=B0j^\vec{B} = B_0\hat{j} exists in the region a<x<2aa < x < 2a and B⃗=−B0j^\vec{B} = -B_0\hat{j}, in the region 2a<x<3a2a < x < 3a, where B0B_0 is a positive constant. A positive point charge moving with a velocity v⃗=v0i^\vec{v} = v_0\hat{i}, where v0v_0 is a positive constant, enters the magnetic field at x=ax = a. The trajectory of the charge in this region can be like,

Diagram 0

Options

A
Option A Diagram 1
Correct
B
Option B Diagram 1
C
Option C Diagram 1
D
Option D Diagram 1

Topics & Concepts

Moving Charges and MagnetismMagnetic Force

Step-by-Step Solution

To find the trajectory of the charged particle, we analyze the magnetic force F⃗\vec{F} acting on it in each region using the Lorentz force equation:
F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B})

Given:

  • Charge q>0q > 0
  • Initial velocity at x=ax = a is v⃗=v0i^\vec{v} = v_0\hat{i}, where v0>0v_0 > 0.

Region 1: a<x<2aa < x < 2a

In this region, the magnetic field is:
B⃗=B0j^(B0>0)\vec{B} = B_0\hat{j} \quad (B_0 > 0)

The magnetic force acting on the particle just as it enters is:
F⃗=q(v0i^×B0j^)=qv0B0k^\vec{F} = q(v_0\hat{i} \times B_0\hat{j}) = q v_0 B_0 \hat{k}

  • Since the force is directed along the positive zz-axis (+k^+\hat{k} direction), the particle accelerates in the +z+z direction.
  • The path is a circular arc in the x-zx\text{-}z plane with its center of curvature located at (a,R)(a, R) in the x-zx\text{-}z plane, where R=mv0qB0R = \frac{mv_0}{qB_0}.
  • As the particle progresses from x=ax = a to x=2ax = 2a, its zz-coordinate increases (z>0z > 0), and the slope dzdx\frac{dz}{dx} increases from 00 (the curve is concave upwards, d2zdx2>0\frac{d^2z}{dx^2} > 0).

Region 2: 2a<x<3a2a < x < 3a

In this region, the magnetic field reverses direction:
B⃗=−B0j^\vec{B} = -B_0\hat{j}

The particle enters this region at x=2ax = 2a with a velocity having components:
v⃗=vxi^+vzk^(vx>0,vz>0)\vec{v} = v_x\hat{i} + v_z\hat{k} \quad (v_x > 0, v_z > 0)

The magnetic force becomes:
F⃗=q[(vxi^+vzk^)×(−B0j^)]=−qvxB0k^+qvzB0i^\vec{F} = q \left[(v_x\hat{i} + v_z\hat{k}) \times (-B_0\hat{j})\right] = -q v_x B_0 \hat{k} + q v_z B_0 \hat{i}

  • The zz-component of the acceleration is now negative:
    az=−qvxB0m<0a_z = -\frac{q v_x B_0}{m} < 0
  • This implies that the curvature of the trajectory reverses, making the curve concave downwards (d2zdx2<0\frac{d^2z}{dx^2} < 0).
  • Since the velocity vector is continuous, the slope dzdx\frac{dz}{dx} remains smooth at x=2ax = 2a (an inflection point), and the particle continues to move in the positive zz-direction while bending towards the horizontal.

Comparing this behavior with the given options, the trajectory starts at (a,0)(a, 0) tangent to the xx-axis, curves upward (concave up) until x=2ax = 2a, and then continues upward with opposite curvature (concave down) until x=3ax = 3a. This corresponds to the curve shown in graph (A).

Correct Answer: (A)

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