Question

A wire carrying a current of $3\text{A}$ is bent in the form of a parabola $y^2=4-x$ as shown in figure, where $x$ and $y$ are in metre. The wire is placed in a uniform magnetic field $\overrightarrow{B}=5\hat{k}\text{T}$. The magnetic force acting on the wire is:

• A. $\left(60\hat{i}\right)\text{N}$
• B. $(-60\hat{i})\text{N}$
• C. $\left(30\hat{i}\right)\text{N}$
• D. $(-30\hat{i})\text{N}$

Answer 1

The correct option is A $\left(60\hat{i}\right)\text{N}$

Joining the initial and final points on the parabolic shaped wire we get the length/displacement vector as shown in figure,


$\overrightarrow{L}=\overrightarrow{NP}$

From the equation of parabola,

$y^2=4-x$

Putting $x=0$ i.e for point $\text{P}$ & $\text{N}$ we have,

$\Rightarrow y^2=4$

or, $y=\pm \sqrt{4}$

$y=\pm 2$

Thus, $\text{|OP|}=\text{|ON|}=2\text{m}$

Hence $\text{NP}=4\text{m}$, $\overrightarrow{L}=\left(4\hat{j}\right)\text{m}$

Using the magnetic force relation we have:

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

${\overrightarrow{F}}_m=3\left[\right(4\hat{j})\times (5\hat{k}\left)\right]$

${\overrightarrow{F}}_m=\left(60\hat{i}\right)\text{N}$