Question:medium

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
  • \( \frac{qdB_0}{m} \)
  • \( \frac{qdB_0}{2m} \)
  • \( \frac{2qdB_0}{m} \)
  • \( \frac{qdB_0}{3m} \)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
A charged particle entering a uniform magnetic field perpendicular to its velocity follows a circular path. To prevent the particle from striking a screen at distance \( d \), the radius of its circular path must be less than or equal to \( d \).
Step 2: Key Formula or Approach:
1. Radius of circular path in magnetic field: \( R = \frac{m V_{\text{total}}}{q B} \).
2. Condition to not hit the screen: \( R \leq d \).
Step 3: Detailed Explanation:
The particle's velocity is given as \( \vec{v}_{\text{particle}} = -2v\hat{i} \). Thus, its speed is \( 2v \).
The magnetic field is \( \vec{B} = B_0\hat{k} \). Since velocity is in the x-direction and field is in the z-direction, they are perpendicular.
The radius of the circular orbit is:
\[ R = \frac{m(2v)}{q B_0} = \frac{2mv}{q B_0} \]
For the particle not to strike the screen at distance \( d \), the radius of its path must satisfy:
\[ R \leq d \]
\[ \frac{2mv}{q B_0} \leq d \]
Solving for \( v \):
\[ v \leq \frac{q d B_0}{2m} \]
Therefore, the maximum value of \( v \) is \( \frac{q d B_0}{2m} \).
Step 4: Final Answer:
The maximum value of \( v \) is \(\frac{q d B_0}{2m}\).
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