Question

A wire lying along $y$ - axis from $y=0$ to $y=1\text{m}$ carries a current of $2\text{mA}$ in the negative $y$ - direction. The wire lies in a non uniform magnetic field given by $\overrightarrow{B}=\left(0.3\text{T/m}\right)y\hat{i}+\left(0.4\text{T/m}\right)y\hat{j}$. The magnetic force on the entire wire will be:

• A. $-3\times {10}^{-4}\hat{j}\text{N}$
• B. $6\times {10}^{-3}\hat{k}\text{N}$
• C. $-3\times {10}^{-4}\hat{k}\text{N}$
• D. $3\times {10}^{-4}\hat{k}\text{N}$

Answer 1

The correct option is D $3\times {10}^{-4}\hat{k}\text{N}$


Here $B_y=0.4y$ and $B_x=0.3y$

As shown in figure the length vector.
$\overrightarrow{L}=(-1\hat{j})\text{m}$

Also,

${\overrightarrow{F}}_m=i(\overrightarrow{L}\times \overrightarrow{B})$ gives the magnetic force experienced by wire.

Due to $y$ - component of magnetic field, the force will be zero, since $\overrightarrow{L}\times \overrightarrow{B}=0$ in this case.

${\overrightarrow{F}}_m=i\left[\right(\overrightarrow{L})\times (0.3y\hat{i}\left)\right]$

$\overrightarrow{F}$ will be variable across the length of wire since $y$ is changing.

Let us assume a small current element of length $dy$, thus $\overrightarrow{dl}=-dy\hat{j}$. Thus mangetic force on this segment will be,

$\overrightarrow{dF}=i(\overrightarrow{dl}\times \overrightarrow{B})$

$\overrightarrow{dF}=i(-dy\hat{j})\times \left(0.3y\hat{i}\right)$

$\overrightarrow{dF}=-0.3iydy(-\hat{k})=\left(0.3iydy\right)\hat{k}$

Thus force on whole wire can be found by integrating and putting the limits $y=0$ to $y=1$

$\overrightarrow{F}={\int }_{y=0}^{y=1}d\overrightarrow{F}$

$\overrightarrow{F}=0.3\times (2\times {10}^{-3}){\int }_0^{1}ydy\hat{k}$

$\overrightarrow{F}=0.6\times {10}^{-3}\times {\left[\frac{y^2}{2}\right]}_0^1\hat{k}$

$\overrightarrow{F}=0.3\times {10}^{-3}\times (1-0)\hat{k}$

$\therefore \overrightarrow{F}=3\times {10}^{-4}\text{N}\hat{k}$