Question

A particle of charge \( q \) moves with a velocity \( \vec{v} = a\hat{i} \) in a magnetic field \( \vec{B} = b\hat{j} + c\hat{k} \), where ' a ', ' b ' and ' c ' are constants. The magnitude of force experienced by particle is

• A. \( qa\sqrt{b^2 + c^2} \)
• B. \( qa(b + c) \)
• C. \( qa\sqrt{b^2 - c^2} \)
• D. zero

Hint:

Force is maximum when velocity is perpendicular to the magnetic field.

Answer 1

The Correct Option is A

Step 1: Lorentz Force Formula
$\vec{F} = q(\vec{v} \times \vec{B})$.
Step 2: Cross Product
$\vec{v} \times \vec{B} = (a\hat{i}) \times (b\hat{j} + c\hat{k}) = a b(\hat{i} \times \hat{j}) + a c(\hat{i} \times \hat{k})$.
Using unit vector rules: $\vec{v} \times \vec{B} = ab\hat{k} - ac\hat{j}$.
Step 3: Magnitude
$|\vec{F}| = q \sqrt{(ab)^2 + (-ac)^2} = q \sqrt{a^2b^2 + a^2c^2} = qa\sqrt{b^2 + c^2}$.
Final Answer: (A)