Question

A wire of length $L$ carries current $I$ along x - axis. A magnetic field $\vec{B} = B_0(\hat{i} - \hat{j} - \hat{k})\text{T}$ acts on the wire. The magnitude of magnetic force acting on the wire is

• A. $\frac{ILB_0}{2}$
• B. $ILB_0$
• C. $2ILB_0$
• D. $\sqrt{2}ILB_0$

Hint:

Only components of B perpendicular to the wire contribute to the force. Here, $B_y$ and $B_z$ contribute.

Answer 1

The Correct Option is D

Step 1: Vector Formula
$\vec{F} = I(\vec{L} \times \vec{B})$.
$\vec{L} = L\hat{i}$.
Step 2: Cross Product
$\vec{F} = I [ L\hat{i} \times B_0(\hat{i} - \hat{j} - \hat{k}) ]$.
$\vec{F} = ILB_0 (\hat{i} \times \hat{i} - \hat{i} \times \hat{j} - \hat{i} \times \hat{k})$.
$\vec{F} = ILB_0 (0 - \hat{k} - (-\hat{j})) = ILB_0 (\hat{j} - \hat{k})$.
Step 3: Magnitude
$|\vec{F}| = ILB_0 \sqrt{1^2 + (-1)^2} = \sqrt{2}ILB_0$.
Final Answer: (D)