Question

In presence of magnetic field $\vec{B}\hat{j}$ and electric field $(-E)\hat{k}$, a particle moves undeflected. Which statement is correct?
 

• A. The particle has positive charge, velocity = $-\dfrac{E}{B}\hat{i}$
• B. The particle has positive charge, velocity = $\dfrac{E}{B}\hat{i}$
• C. The particle has negative charge, velocity = $-\dfrac{E}{B}\hat{i}$
• D. The particle has negative charge, velocity = $\dfrac{E}{B}\hat{i}$

Hint:

For undeflected motion in crossed E and B fields, $v = \dfrac{E}{B}$.

Answer 1

The Correct Option is B, D

Step 1: Condition for undeflected motion.
Lorentz force must vanish: \[ q(\vec{E} + \vec{v} \times \vec{B}) = 0 \]

Step 2: Substitute fields.
$\vec{E} = -E\hat{k}$, $\vec{B} = B\hat{j}$. Let velocity = $v\hat{i}$. Compute cross product: \[ \vec{v} \times \vec{B} = v\hat{i} \times B\hat{j} = vB\hat{k} \] Force condition: \[ -E\hat{k} + vB\hat{k} = 0 $\Rightarrow$ vB = E $\Rightarrow$ v = \frac{E}{B} \]

Step 3: Charge sign.
Force cancels only if $q>0$. Thus particle is positive and moves along +x direction.

Step 4: Conclusion.
Correct answer is (B).