Magnetic Field Interface Problem — Smart Geometry Method 💡

❓ 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 then continues to move into region 1 and finally reaches the interface again.

What is the displacement of the particle along the interface during this complete motion?

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🖼️ Question Image

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✍️ Short Solution

This is a pure concept-based JEE magnetism problem.
No equations of motion — only circular motion + geometry.

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🔹 Step 1 — Motion of a charged particle in uniform magnetic field

When a charged particle with velocity $v$ enters a region with a uniform magnetic field $B$ (perpendicular to the velocity):

• The particle moves in a circular path

• Radius of circular motion:

$$r=\frac{mv}{qB}$$

So in the two regions:

$$r_1=\frac{mv}{q{B}_1},r_2=\frac{mv}{q{B}_2}$$

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🔹 Step 2 — What happens at the interface?

• Particle starts on the interface

• Enters region 2 → moves along a circular arc of radius $r_2$

• Returns to the interface

• Then enters region 1 → moves along a circular arc of radius $r_1$

• Finally reaches the interface again

📌 Important:
The speed remains constant (magnetic force does no work).

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🔹 Step 3 — Key geometric observation (MOST IMPORTANT 🔥)**

In each region, the particle completes a semicircular turn before returning to the interface.

• Displacement along interface due to motion in region 2:

$$\Delta x_2 = 2r_2$$

• Displacement along interface due to motion in region 1:

$$\Delta x_1 = 2r_1$$

But these displacements are in opposite directions.

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🔹 Step 4 — Net displacement along the interface

So total displacement along the interface:

$$\Delta x = 2r_2 - 2r_1$$

Substitute radii:

$$\Delta x = 2\left(\frac{mv}{qB_2}-\frac{mv}{qB_1}\right)$$

Factor out constants:

$$\boxed{ \Delta x = \frac{2mv}{q} \left(\frac{1}{B_2}-\frac{1}{B_1}\right) }$$

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🔹 Step 5 — Direction insight (JEE favourite 🧠)**

• If $B_2 < B_1$ → $r_2 > r_1$ → net displacement in direction of region 2 arc

• If $B_2 > B_1$ → displacement reverses direction

📌 JEE usually asks magnitude, not direction.

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✅ Final Answer

$$\boxed{ \Delta x = \frac{2mv}{q} \left(\frac{1}{B_2}-\frac{1}{B_1}\right) }$$