Question:
A particle of mass m and charge q has an initial velocity \( \overrightarrow v = {v_0}\widehat j \) . If an electric field \( \overrightarrow E = {E_0}\widehat i \) and magnetic field \( \overrightarrow B = {B_0}\widehat i \) act on the particle, its speed will double after a time:
A particle of mass m and charge q has an initial velocity \( \overrightarrow v = {v_0}\widehat j \) . If an electric field \( \overrightarrow E = {E_0}\widehat i \) and magnetic field \( \overrightarrow B = {B_0}\widehat i \) act on the particle, its speed will double after a time:
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Magnetic force does no work and, with \(\vec B\parallel\hat i\), only rotates the velocity components in the y-z plane. The x-motion is due to E alone: \(v_x=qE_0t/m\).
Updated On: Oct 31, 2026
- \( {{3m{v_0}} \over {q{E_0}}} \)
- \( {{\sqrt 2 m{v_0}} \over {q{E_0}}} \)
- \( {{\sqrt 3 m{v_0}} \over {q{E_0}}} \)
- \( {{2m{v_0}} \over {q{E_0}}} \)
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The Correct Option is C
Solution and Explanation
Step 1: Magnetic force \(q\vec v\times\vec B\) is perpendicular to \(\vec B\parallel\hat i\), so it only affects \(v_y,v_z\); it changes their direction but not the magnitude \(\sqrt{v_y^2+v_z^2}=v_0\).
Step 2: Along x only the electric force acts: \(a_x=\dfrac{qE_0}{m}\), \(v_x=\dfrac{qE_0}{m}t\).
Step 3: Speed\(^2=v_0^2+\left(\dfrac{qE_0t}{m}\right)^2=(2v_0)^2\Rightarrow\dfrac{qE_0t}{m}=\sqrt3\,v_0\).
Step 4: \(t=\dfrac{\sqrt3\,m v_0}{qE_0}\).
Answer: (C) \(\dfrac{\sqrt3\,m v_0}{qE_0}\)
Step 2: Along x only the electric force acts: \(a_x=\dfrac{qE_0}{m}\), \(v_x=\dfrac{qE_0}{m}t\).
Step 3: Speed\(^2=v_0^2+\left(\dfrac{qE_0t}{m}\right)^2=(2v_0)^2\Rightarrow\dfrac{qE_0t}{m}=\sqrt3\,v_0\).
Step 4: \(t=\dfrac{\sqrt3\,m v_0}{qE_0}\).
Answer: (C) \(\dfrac{\sqrt3\,m v_0}{qE_0}\)
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