Question

A particle of charge \(1.0 \times 10^{-16}\) C moves through a uniform magnetic field \(\vec{B} = B_0 (\hat{i} + 4\hat{j})\) T. The particle velocity at some instant is \(\vec{V} = (2\hat{i} + 4\hat{j})\) m/s and the magnetic force acting on it is \(3 \times 10^{-16}\) N. The magnitude of \(B_0\) is

• A. 1.0 T
• B. 2.5 T
• C. 0.5 T
• D. 0.75 T

Hint:

Use cross product for vector magnetic force.
Only perpendicular components contribute to force.
Compute magnitude of force from \(|q||\vec{v} \times \vec{B}|\).

Answer 1

The Correct Option is D

The magnetic force on a charged particle is given by: \[ \vec{F} = q (\vec{v} \times \vec{B}) \] Given: \[ q = 1 \times 10^{-16} \text{ C}, \quad \vec{v} = 2\hat{i} + 4\hat{j}, \quad \vec{B} = B_0(\hat{i} + 4\hat{j}) \] Now calculate: \[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 0 \\ B_0 & 4B_0 & 0 \\ \end{vmatrix} = (0)\hat{i} - (0)\hat{j} + (2 \cdot 4B_0 - 4 \cdot B_0)\hat{k} = (8B_0 - 4B_0)\hat{k} = 4B_0 \hat{k} \] So, \[ |\vec{F}| = q \cdot |4B_0| = 1 \times 10^{-16} \cdot 4B_0 = 3 \times 10^{-16} \Rightarrow B_0 = \frac{3 \times 10^{-16}}{4 \times 10^{-16}} = 0.75 \text{ T} \]