Question

A conducting circular loop is placed in a uniform magnetic field $B=0.025T$ with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at $1mm{{s}^{-1}}$. The induced emf in the loop when the radius is 2 cm is:
A. $2\pi \mu V$
B. $\pi \mu V$
C. $\dfrac{\pi }{2}\mu V$
D. $3.2\pi \mu V$

Answer 1

Use the formula for the induced emf due to a changing magnetic flux through a loop. Write an expression for the magnetic flux through the given loop in terms of the radius and differentiate it with respect to the radius. Then use the given data to calculate the induced emf when the radius is 2cm.

Formula used:
$E=\dfrac{d\phi }{dt}$
$\phi =\overrightarrow{B}.\overrightarrow{A}=BA\cos \theta$

Complete step by step answer:
When magnetic flux through a conducting loop changes with time, an emf is induced in the loop. As a result a current flows in the loop. The emf induced in the loop due to varying magnetic flux is given as $E=\dfrac{d\phi }{dt}$, where E is the emf induced, $\phi$ is the magnetic flux and t is time. We can also say that the induced emf is equal to the rate of change in the magnetic flux through the loop.

The magnetic flux through a loop of constant (uniform) magnetic field is given as
$\phi =\overrightarrow{B}.\overrightarrow{A}=BA\cos \theta$ …. (i).
Here, B is the magnitude of the magnetic field, A is the area enclosed by the loop and $\theta$ is the angle between the magnetic field and the area vector.

It is said that that magnetic field is perpendicular to the plane of the loop. This means that the magnetic field and the area vector of the loop are parallel. Therefore, $\theta =0$.
And $\cos 0=1$.

Also, it is given that $B=0.025T$ and the area enclosed by the loop will be equal to $A=\pi {{r}^{2}}$ (r is the radius of the loop).
Substitute the value in (i).
$\Rightarrow \phi =(0.025)\left( \pi {{r}^{2}} \right)$ ….. (ii).

Now differentiate (ii) with respect to r.
$\dfrac{d\phi }{dt}=\dfrac{d}{dt}\left[ (0.025)\left( \pi {{r}^{2}} \right) \right]$
$\Rightarrow \dfrac{d\phi }{dt}=0.05\pi r\dfrac{dr}{dt}$.
It is given that the radius of the loop is shrinking at rate of $1mm{{s}^{-1}}={{10}^{-3}}m{{s}^{-1}}$.
This means that $\dfrac{dr}{dt}={{10}^{-3}}m{{s}^{-1}}$.
And we know that $E=\dfrac{d\phi }{dt}$.
$\Rightarrow E=0.05\pi r\dfrac{dr}{dt}$.
Substitute the value of $\dfrac{dr}{dt}$.
$E=0.05\pi r\left( {{10}^{-3}} \right)\\\ \Rightarrow E =5\times {{10}^{-5}}\pi r$.
Now, substitute $r=2cm=0.02m$ in the above equation.
$E={{10}^{-5}}\pi (0.02)\\\ \Rightarrow E ={{10}^{-6}}\pi \\\ \therefore E =\pi \mu V$.
This means that the emf induced in the loop when the radius of the loop is 2 cm is $\pi \mu V$.

Hence, the correct option is B.

Note: In some books you find that the emf induced in a loop due to the varying magnetic flux is equal to $E=-\dfrac{d\phi }{dt}$.
Here, the negative sign only indicates that the emf is induced in such a way that it opposes the change in current in the loop. It is advisable not to use the negative sign as it may confuse things.