Question

A particle of charge \(1.0\times10^{-16}\ \text{C}\) moves through a uniform magnetic field \[ \vec{B}=B_0(\hat{i}+4\hat{j})\ \text{T}. \] The particle velocity at some instant is \[ \vec{V}=(2\hat{i}+4\hat{j})\ \text{m s}^{-1} \] and the magnetic force acting on it is \[ 3\times10^{-16}\hat{k}\ \text{N}. \] The magnitude of \(B_0\) is

• A. \(1.0\ \text{T}\)
• B. \(2.5\ \text{T}\)
• C. \(0.5\ \text{T}\)
• D. \(0.75\ \text{T}\)

Hint:

For magnetic force problems, always use \[ \vec{F}=q(\vec{V}\times\vec{B}). \] When vectors are given in component form, evaluate the cross product using a determinant. The direction is obtained from the right-hand rule.

Answer 1

The Correct Option is D

Step 1: Compute \( \vec{V}\times\vec{B} \).
\( \vec{V} = 2\hat{i}+4\hat{j} \), \( \vec{B} = B_0(\hat{i}+4\hat{j}) \). \[ \vec{V}\times\vec{B} = B_0[(4\cdot0-0\cdot4)\hat{i}-(2\cdot0-0\cdot1)\hat{j}+(2\cdot4-4\cdot1)\hat{k}] = 4B_0\hat{k} \]

Step 2: Equate to the given force.
\( F = q(\vec{V}\times\vec{B}) \Rightarrow 3\times10^{-16} = 1.0\times10^{-16}\times4B_0 \Rightarrow B_0 = 0.75\,\text{T} \). \[ \boxed{0.75\,\text{T}} \]