A small square loop of wire of side $$\ell $$ is placed inside a large square loop of wire of side $$L\left( {L > \ell } \right).$$ The loop are coplanar and their centre coincide. The mutual inductance of the system is proportional to
A.
$$\frac{\ell }{L}$$
B.
$$\frac{{{\ell ^2}}}{L}$$
C.
$$\frac{L}{\ell }$$
D.
$$\frac{{{L^2}}}{\ell }$$
Answer :
$$\frac{{{\ell ^2}}}{L}$$
Solution :
$${\phi _{{\text{total}}}} = {B_{{\text{large}}}}{A_{{\text{small}}}} = \frac{{{\mu _0}}}{{4\pi }}\frac{i}{{\frac{L}{2}}}\left( {2\sin {{45}^ \circ }} \right) \times {\ell ^2}$$
On comparing with $${\phi _{{\text{total}}}} = Mi,$$ we get $$M \propto \frac{{{\ell ^2}}}{L}$$
Releted MCQ Question on Electrostatics and Magnetism >> Electromagnetic Induction
Releted Question 1
A thin circular ring of area $$A$$ is held perpendicular to a
uniform magnetic field of induction $$B.$$ $$A$$ small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is $$R.$$ When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer is
A thin semi-circular conducting ring of radius $$R$$ is falling with its plane vertical in horizontal magnetic induction $$\overrightarrow B .$$ At the position $$MNQ$$ the speed of the ring is $$v,$$ and the potential difference developed across the ring is
A.
zero
B.
$$\frac{{Bv\pi {R^2}}}{2}$$ and $$M$$ is at higher potential
Two identical circular loops of metal wire are lying on a table without touching each other. Loop-$$A$$ carries a current which increases with time. In response, the loop-$$B$$
A coil of inductance $$8.4 mH$$ and resistance $$6\,\Omega $$ is connected to a $$12 V$$ battery. The current in the coil is $$1.0 A$$ at approximately the time