Question

A wire of length $l$ carries a current $I$ along $\text{x-}$axis. A magnetic field exists which is given as $\overrightarrow{B}=B_0(\hat{i}+\hat{j}+\hat{k})\text{T}$. The magnitude of the magnetic force acting on the wire will be

• A. $B_{0}Il$
• B. $B_{0}Il\sqrt{2}$
• C. $2B_{0}Il$
• D. $\frac{B_{0}Il}{\sqrt{2}}$

Answer 1

The correct option is B $B_{0}Il\sqrt{2}$

The wire is carrying current along the positive $x-$axis, so$\overrightarrow{L}=l\hat{i}$

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

Magnetic force on current carrying wire will be,

$\overrightarrow{F}=(I\overrightarrow{L}\times \overrightarrow{B})$

$\Rightarrow \overrightarrow{F}=[Il\hat{i}\times B_0(\hat{i}+\hat{j}+\hat{k}\left)\right]$

$\Rightarrow \overrightarrow{F}=Il{B}_0\left[\right(\hat{i}\times \hat{i})+(\hat{i}\times \hat{j})+(\hat{i}\times \hat{k}\left)\right]$

$\Rightarrow \overrightarrow{F}=Il{B}_0[0+\hat{k}-\hat{j}]$

$\Rightarrow \overrightarrow{F}=-B_{0}Il\hat{j}+B_{0}Il\hat{k}$

$\therefore |\overrightarrow{F}|=\sqrt{{\left(B_{0}Il\right)}^2+{\left(B_{0}Il\right)}^2}=\sqrt{2}{B}_{0}Il$

Hence, option (b) is the correct answer.