Question:
A particle of charge \( q \) and mass \( m \) is moving with a velocity \( -2v \hat{i} \) (\( v \neq 0 \)) towards a large screen placed in YZ-plane placed at a distance \( d \). If there is a magnetic field \( \vec{B} = B_0 \hat{k} \), the maximum value of \( v \) for which the particle will not strike the screen is
A particle of charge \( q \) and mass \( m \) is moving with a velocity \( -2v \hat{i} \) (\( v \neq 0 \)) towards a large screen placed in YZ-plane placed at a distance \( d \). If there is a magnetic field \( \vec{B} = B_0 \hat{k} \), the maximum value of \( v \) for which the particle will not strike the screen is
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If particle must not reach a boundary $\longrightarrow$ compare radius of circular path with given distance.
Updated On: Apr 22, 2026
- \( \frac{qdB_0}{m} \)
- \( \frac{qdB_0}{2m} \)
- \( \frac{2qdB_0}{m} \)
- \( \frac{qdB_0}{3m} \)
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The Correct Option is B
Solution and Explanation
Concept:
Charged particle in magnetic field moves in circular path:
\[
r = \frac{mv}{qB}
\]
Step 1: Understand motion.
Velocity along \( -x \), magnetic field along \( z \) \(\Rightarrow\) circular motion in XY-plane.
Step 2: Condition to avoid screen.
Particle should not reach distance \( d \) along x-axis. Maximum displacement in circular motion = radius. \[ r \le d \]
Step 3: Substitute radius.
\[ \frac{m(2v)}{qB_0} \le d \Rightarrow v \le \frac{qdB_0}{2m} \]
Step 1: Understand motion.
Velocity along \( -x \), magnetic field along \( z \) \(\Rightarrow\) circular motion in XY-plane.
Step 2: Condition to avoid screen.
Particle should not reach distance \( d \) along x-axis. Maximum displacement in circular motion = radius. \[ r \le d \]
Step 3: Substitute radius.
\[ \frac{m(2v)}{qB_0} \le d \Rightarrow v \le \frac{qdB_0}{2m} \]
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