Question

Two conducting circular loops of radii ${R_1}$​ and ${R_2}$ (${R_1} > > {R_2}$​) are placed in the same plane with their centers coinciding. Find the mutual inductance between them.
(A) $\dfrac{{{\mu _0}\pi {R_1}^2}}{{{R_2}}}$
(B) $\dfrac{{{\mu _0}\pi {R_2}^2}}{{{R_1}}}$
(C) $\dfrac{{{\mu _0}\pi {R_1}^2}}{{2{R_2}}}$
(D) $\dfrac{{{\mu _0}\pi {R_2}^2}}{{2{R_1}}}$

Answer 1

Hint It is given that there are two conducting circular loops of specified radius, and given that radius of the first loop is greater than the second. Now, it is understandable that the second loop is the inner loop. Using flux density of the magnetic field formula, find the resulting flux.

Complete Step By Step Answer
It is given that there are two closed conducting loops of radius ${R_1}$​ and ${R_2}$and where ${R_1}$is said to be very larger than ${R_2}$. From this statement we can assume that that circle with radius ${R_2}$is the inner circle, circumscribed by the circle with radius ${R_1}$. Now, let us assume that there is a current flowing through the bigger circle coil. Now, the magnetic field at the center of the coil is given by the ratio of current flowing through the conductor and the diameter of the coil. Mathematically,
$\Rightarrow B = \dfrac{{{\mu _0}i}}{{2{R_1}}}$, where $i$ is current flowing through the coil and${\mu _0}$ is permeability of free space.
Since both the circles share a similar center, there will be direct influence on the inner coil due to the magnetic field of the first one. Now, this causes change in flux in the second coil, which is mathematically given as the product of magnetic field at the center and the area of the circle where a flux is induced.
$\Rightarrow \phi = B \times A$
Flux is induced in inner coil of radius ${R_2}$, substituting for B and area, we get
$\Rightarrow \phi = \dfrac{{{\mu _0}i}}{{2{R_1}}} \times \pi {R_2}^2$
Now, mutual inductance is given using the formula,
$\Rightarrow \phi = Mi$(where M is the mutual inductance and $i$ is current flowing through the coil)
$\Rightarrow \dfrac{{{\mu _0}i\pi {R_2}^2}}{{2{R_1}}} = Mi$
Cancelling out common term we get
$\Rightarrow \dfrac{{{\mu _0}\pi {R_2}^2}}{{2{R_1}}} = M$

Thus, option(d) is the right answer for a given question.

Note Mutual Inductance is defined as the phenomenon where the magnetic field caused due to the current flowing through one coil, induces an EMF on the nearby adjacent coil, when it comes in contact with the magnetic field of the first coil.