Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 7 April, Morning Shift — Question 57

Uniform magnetic fields of different strengths ( B1B_{1} and B2B_{2} ), both normal to the plane of the paper exist as shown in the figure. A charged particle of mass mm and charge qq, at the interface at an instant, moves into the region 2 with velocity vv 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?

figure

  1. Option A:

    mvqB1(1−B1B2)\frac{m v}{q B_{1}}\left(1-\frac{B_{1}}{B_{2}}\right)

  2. Option B:

    mvqB1(1−B2B1)\frac{m v}{q B_{1}}\left(1-\frac{B_{2}}{B_{1}}\right)

  3. Option C:

    mvqB1(1−B2B1)×2\frac{m v}{q B_{1}}\left(1-\frac{B_{2}}{B_{1}}\right) \times 2

  4. Option D:

    mvqB1(1−B1B2)×2\frac{m v}{q B_{1}}\left(1-\frac{B_{1}}{B_{2}}\right) \times 2

    Correct

Answer: D

Step-by-step solution

R1=mvqB1R_{1}=\frac{m v}{q B_{1}}

R2=mvqB2R_{2}=\frac{m v}{q B_{2}}

∣Δr∣=∣2R2−2R1∣=2mvq[1B2−1B1]|\Delta r|=\left|2 R_{2}-2 R_{1}\right|=\frac{2 m v}{q}\left[\frac{1}{B_{2}}-\frac{1}{B_{1}}\right]

⇒Δr=∣2mvqB1[B1B2−1]∣=∣2mvqB1(1−B1B2)∣\Rightarrow \quad \Delta r=\left|\frac{2 m v}{q B_{1}}\left[\frac{B_{1}}{B_{2}}-1\right]\right|=\left|\frac{2 m v}{q B_{1}}\left(1-\frac{B_{1}}{B_{2}}\right)\right|

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Motion of Charged Particles in magnetic Fields