Question

A square loop of side $l$ enters a region of transverse magnetic field with a constant speed $v$ as shown in the figure. The resistance of the loop is $r$. Choose the graph that represents the variation of induced current $\left(i\right)$ across the loop versus time. [Given, $d>2l$]

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

As, given side of the square loop is $l$ and at time $t=0$ the loop is in the verge of entering into the field region $d$. Thus, the time taken by the loop to completely enter into the field region is,

$t=\frac{l}{v}$

Case $\left(\text{I}\right):$

For the time interval $[0<t<\frac{l}{v}]$,

The effective area of the loop in the field region is $A_1=lvt$

Flux linked with the loop is, $\varphi =Blvt$

$\therefore \left|\begin{array}{c}E\end{array}\right|=\left|\begin{array}{c}-\frac{d\varphi }{dt}\end{array}\right|=Blv$

Hence current in the loop is, $i=\frac{E}{r}=\frac{Blv}{r}$


Case $\left(\text{II}\right):$

For the time interval $[\frac{l}{v}<t<\frac{d}{v}]$,

In this interval, the given loop completely remains inside the magnetic field. Since the field is uniform, flux linked with the loop is,

$\varphi =B{l}^2=\text{constant}$

$\therefore \left|\begin{array}{c}E\end{array}\right|=\left|\begin{array}{c}-\frac{d\varphi }{dt}\end{array}\right|=0$

Hence, current in the loop is zero i.e., $i=\frac{E}{R}=0$

Case $\left(\text{III}\right):$

Now, for the time interval $[\frac{d}{v}<t<\frac{d+l}{v}]$,

In this interval, the given loop comes out from the magnetic field. Since the field is uniform. Thus flux linked with coil starts decreasing gradually.

Hence, we will get similar result as case $\left(\text{I}\right)$, but the direction of induced current would be opposite to that of case $\left(\text{I}\right)$, as shown in the figure.


Hence, option $\left(\text{D}\right)$ is correct.