JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Magnetic Effects of Current

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 the 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) $\frac{m v}{q B _{1}} (1 - \frac{B _{2}}{B _{1}}) \times 2$
(B) $\frac{m v}{q B _{1}} (1 - \frac{B _{1}}{B _{2}})$
(C) $\frac{m v}{q B _{1}} (1 - \frac{B _{2}}{B _{1}})$
(D) $\frac{m v}{q B _{1}} (1 - \frac{B _{1}}{B _{2}}) \times 2$ ✓ Correct

Answer: (D)

Solution

In a uniform field the particle moves on a circle of radius

R = \frac{m v}{q B}

, and after dipping across the interface it re-meets the interface displaced by one diameter. In region 2 the shift is

2 R_{2} = \frac{2 m v}{q B _{2}}

, and in region 1 it is

2 R_{1} = \frac{2 m v}{q B _{1}}

. The particle curves one way below the interface and the opposite way above it, so the two shifts oppose and the net displacement along the interface is

2 R_{1} - 2 R_{2} = \frac{2 m v}{q} (\frac{1}{B _{1}} - \frac{1}{B _{2}}) = \frac{2 m v}{q B _{1}} (1 - \frac{B _{1}}{B _{2}}) .

The displacement along the interface is

\frac{m v}{q B _{1}} (1 - \frac{B _{1}}{B _{2}}) \times 2

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Magnetic Effects of Current

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