Question

A proton moves at a constant speed \(v\) along the x-axis through a crossed electric and magnetic field. The magnetic field is \(\vec{B}=B_{0}\hat{k}\) T. The electric field \(\vec{E}\) in V/m is

• A. \(-vB_{0}\hat{j}\)
• B. \(vB_{0}\hat{j}\)
• C. \(vB_{0}\hat{k}\)
• D. \(vB_{0}\hat{i}\)

Hint:

For constant velocity the electric and magnetic forces must cancel, so \(\vec{E}=-\vec{v}\times\vec{B}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A proton moving at a constant velocity has zero net force: \(q(\vec{E}+\vec{v}\times\vec{B})=0\), so \(\vec{E}=-\vec{v}\times\vec{B}\).

Step 2: Calculation
With \(\vec{v}=v\hat{i}\) and \(\vec{B}=B_0\hat{k}\):
\(\vec{v}\times\vec{B}=vB_0(\hat{i}\times\hat{k})=-vB_0\hat{j}\).

Step 3: Electric field
\(\vec{E}=-(-vB_0\hat{j})=vB_0\hat{j}\).
The electric force on the proton is along \(+\hat{j}\) and cancels the magnetic force along \(-\hat{j}\).

Final Answer:
The electric field is \(vB_0\hat{j}\) V/m. \[ \boxed{\text{(B) }vB_{0}\hat{j}} \]