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, $\overrightarrow{B}=(0.3y\hat{i}+0.4y\hat{j})\text{T}$. The magnetic force on the entire wire will be -

• A. $(-6\times {10}^{-3}\hat{k})\text{N}$
• 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 $(3\times {10}^{-4}\hat{k})\text{N}$



The wire is lying along $y-$axis, and the direction of the current is along the negative $y-$axis.

Let us consider an element of length $dy$, at a distance $y$, from the origin and on the $y$ axis.

$\therefore \overrightarrow{dl}=-dy\hat{j}$

Magnetic force on the current carrying conductor element is,

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

Now, magnetic force on the entire conductor,

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

$\therefore \overrightarrow{F}=-i{\int }_0^l[dy\hat{j}\times (0.3\hat{i}+0.4\hat{j}\left)y\right]$


$\Rightarrow \overrightarrow{F}=-i{\int }_0^1(-0.3ydy)\hat{k}$

$=2\times {10}^{-3}\times 0.3\times {\left[\frac{y^2}{2}\right]}_0^1\hat{k}$

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(D\right)$ is the correct answer.

Why this question? This question has been asked to make you understand the concept of force experienced by a current carrying conductor in a magnetic field.