Question

A plane loop shown in the figure is shaped in the form of two squares with sides $a\text{and}b$ is introduced into a uniform magnetic field at right angles, to the loop's plane. The magnetic induction varies with time as $B=B_0\sin \omega t$. Find the amplitude of the current induced in the loop if its resistance per unit length is equal to $r$. The induction of the loop is negligible.

• A. $\frac{(a-b)B_0\omega }{r}$
• B. $\frac{1}{2}\frac{(a-b)B_0\omega }{r}$
• C. $\frac{1}{4}\frac{(a-b)B_0\omega }{r}$
• D. $\frac{2(a-b)B_0\omega }{r}$

Answer 1

The correct option is C $\frac{1}{4}\frac{(a-b)B_0\omega }{r}$

The loops are connected in such a way that area vector will oppose each other.

Assuming the loop of side $a$ parallel to the magnetic field $B$, the net flux through the given loops will be

${\varphi }_{net}={\varphi }_1+{\varphi }_2=B{a}^2\cos 0^{\circ }+B{b}^2\cos {180}^{\circ }$

$=B(a^2-b^2)$

So, the magnitude of net $\text{emf}$ induced in the loops is,

$E_{net}=\frac{d{\varphi }_{net}}{dt}=\frac{d}{dt}(a^2-b^2)B$

$=(a^2-b^2)\frac{d}{dt}\left(B_0\sin \omega t\right)=(a^2-b^2)B_0\omega \cos \omega t$

Resistance of the circuit is given as,

$R=4(a+b)r$

The amplitude of current induced be,

$i_0=\frac{E_0}{R}$

As, amplitude of the induced $\text{emf}$ is,

$E_0=(a^2-b^2)B_0\omega$

$\Rightarrow i_0=\frac{(a^2-b^2)B_0\omega }{4(a+b)r}$

$\therefore i_0=\frac{1}{4}\frac{(a-b)B_0\omega }{r}$

Hence, option $\left(C\right)$ is correct.