Question

A small square loop of wire whose side is $l$ which is inside another large square loop of side $L(L > > l).$ Both loops are coplanar and their centres coincide. The mutual inductance of the system is proportional to
A. $\dfrac{l}{L}$
B. $\dfrac{{{l^2}}}{L}$
C. $\dfrac{L}{l}$
D. $\dfrac{{{L^2}}}{l}$

Answer 1

Hint:Calculate the expression of magnetic field in the centre of square from the expression-
$B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{i \times 2\sin {{45}^ \circ }}}{{\dfrac{L}{2}}}$
Then, calculate the magnetic flux of the smaller loop by using-
$\phi = \vec B.\vec S$
Now, mutual inductance is calculated by-
$M = \dfrac{\phi }{i}$

Complete step-by-step solution:
Electromagnetic induction is the process or phenomenon to generate the electric effect by using magnetic effect.
The number of magnetic field lines passing perpendicular to the surface is called magnetic flux. It is denoted by $\phi .$ The expression of magnetic flux is-
$
  \phi = \vec B.\vec S \\
  \phi = BS\cos \theta \\
 $
The first law of Faraday’s law of electromagnetic radiation states that whenever magnetic flux passing through surface change with respect to time there will be an induced EMF generated.
Now, according to the question we will calculate the mutual induction of the system.
Let the length of larger square and smaller square be $L$ and $l$ respectively and current flowing in larger square and small square be $I$ and $i$ respectively.
We know that, magnetic field at a distance $\dfrac{L}{2}$ from the centre of current carrying wire of length $L$ is –
$B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{i}{{\dfrac{L}{2}}}2\sin {45^ \circ }$
Magnetic field in the centre of square due to four wires of length $L - $
$
  {B_1} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{4 \times 2 \times i \times 2 \times \dfrac{1}{{\sqrt 2 }}}}{L} \\
  {B_1} = \dfrac{{{\mu _0}}}{\pi }\dfrac{{2\sqrt 2 i}}{L} \\
 $
As the smaller loop is at the centre and $L > > l$.
Magnetic field in a small square is uniform and is equal to ${B_1}$.
Therefore, the flux of small loop is –
$\phi = {B_1}{l^2}$
$\phi = \dfrac{{{\mu _0}}}{\pi }\dfrac{{2\sqrt 2 i{l^2}}}{L}$
Now, mutual inductance is given by –
$
  M = \dfrac{\phi }{i} \\
  M = \dfrac{{\dfrac{{{\mu _0}}}{\pi }\dfrac{{2\sqrt 2 i{l^2}}}{L}}}{i} \\
 $
Cancelling $i$ in numerator and denominator, we get
$
  M = \dfrac{{{\mu _0}}}{\pi }\dfrac{{2\sqrt 2 {l^2}}}{L} \\
  M\infty \dfrac{{{l^2}}}{L} \\
 $
Therefore, the correct option is (B).

Note:- Mutual inductance can be defined as inductance in which the EMF is induced in the coil due to changing of current of another coil. It depends upon the number of turns in the neighboring coil, area of cross – section and closeness of two coils.