Question

A wire of length $l$ carries a current $i$ along x-axis. A magnetic field exits given by $B=B_0(\hat{i}+\hat{j}+\hat{k})T$. The magnitude of the magnetic force acting on the wire is

• A. $il{B}_0$
• B. $\sqrt{3}{B}_{0}il$
• C. $2il{B}_0$
• D. $\sqrt{2}{B}_{0}il$

Answer 1

The correct option is D $\sqrt{2}{B}_{0}il$

Length = I, Current = $I\hat{i}$

$\overrightarrow{B}=B_0(\hat{i}+\hat{j}+\hat{k})T=B_0\hat{i}+B_0\hat{j}+B_0\hat{k}T$

$F=I|\times \overrightarrow{B}=I|\hat{i}\times B_0\overrightarrow{i}+B_0\hat{j}+B_0\hat{k}$

$=I|B_0\hat{i}\times \hat{i}+|B_0\hat{i}\times \hat{j}+|B_0\hat{i}\times \hat{k}=I|B_0\hat{k}-I|B_0\hat{j}$

or, $|\overrightarrow{F}|=\sqrt{2I^2{I}^2{B}_0^2}=\sqrt{2}I|B_0$