Question

Uniform magnetic fields of different strengths $ B_1 $ and $ B_2 $, both normal to the plane of the paper, exist as shown in the figure. A charged particle of mass $ m $ and charge $ q $, at the interface at an instant, moves into region 2 with velocity $ v $ and returns to the interface. It continues to move into region 1 and finally reaches the interface. What is the displacement of the particle during this movement along the interface?

Consider the velocity of the particle to be normal to the magnetic field and  $ B_2 > B_1 $.

• A. \( \frac{mv}{qB_1} \left( 1 - \frac{B_2}{B_1} \right) \times 2 \)
• B. \( \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \)
• C. \( \frac{mv}{qB_1} \left( 1 - \frac{B_2}{B_1} \right) \)
• D. \( \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2 \)

Hint:

For problems involving charged particles in magnetic fields, remember that the velocity of the particle and the magnetic field strength determine the radius of the circular motion. Use this to relate the displacement.

Answer 1

The Correct Option is D

We are given that a charged particle moves from one region to another and returns back, passing through an interface between two regions with different magnetic field strengths. 
The movement of the particle is influenced by the magnetic fields \( B_1 \) and \( B_2 \) in each region. 
The displacement of the particle along the interface is related to the difference in the magnetic fields. 
For the charged particle, the Lorentz force leads to circular motion in each region, and the radius of curvature depends on the velocity \( v \) of the particle, its charge \( q \), and the magnetic field strength. 
The displacement is related to the difference between the magnetic fields and can be expressed by the formula: \[ \text{Displacement} = \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2. \] 
Thus, the correct answer is: \[ \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2. \]

Answer 2

As \(\vec{v} \perp \vec{B}\), the charged particle will move in a **circular path**, whose radius is given by: \[ R = \frac{mv}{qB} \] Starting point → \(A\) Ending point → \(C\) Hence, the **net displacement** is: \[ \text{AC} = \text{CD} - \text{AD} \] Now, \[ AC = \frac{2mv}{qB_1} - \frac{2mv}{qB_2} \] Simplifying, \[ AC = \frac{2mv}{qB_1} \left(1 - \frac{B_1}{B_2}\right) \] \[ \boxed{AC = \frac{2mv}{qB_1} \left(1 - \frac{B_1}{B_2}\right)} \]