Question 245: Electromagnetic Induction

Question

The planar concentric rings of metal wire having radii $r_1$ and $r_2$ (with $r_1>r_2$ ) are placed in air. The current $I$ is flowing through the coil of larger radius. The mutual inductance between the coils is given by ( $\mu_0=$ permeability of free space)

Accepted Answer

The magnetic field at the centre of a loop is given by
$$\mathrm{B}=\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}\quad	ext{.... (i)}$$
$	herefore \quad$ Magnetic field produced by ring $A, B_A=\frac{\mu_0 I}{2 r_1}$
$	herefore \quad$ Magnetic flux produced in ring B due to $\mathrm{B}_{\mathrm{A}}$,
$$\begin{array}{lll}
	herefore & \phi_B=B_A A_B & \ldots\left(A_B 	ext { is area }ight) \
	herefore & \phi_B=\frac{\pi_0 I}{2 r_1} 	imes \pi r_2^2=\frac{\mu_0 \pi r_2^2}{2 r_1} I & \ldots[	ext { From(i})]
\end{array}$$
Mutual Inductance $M=\frac{\phi}{I}$
$$	herefore \quad M=\frac{\phi_{\mathrm{B}}}{\mathrm{I}}=\frac{\mu_0 \pi r_2^2 \mathrm{I}}{2 \mathrm{r}_1 \mathrm{I}}=\frac{\mu_0 \pi r_2^2}{2 r_1}$$

Answer

$\frac{\mu_0 \pi\left(r_1+r_2\right)^2}{2 r_2}$

Answer

$\frac{\mu_0 \pi\left(r_1-r_2\right)^2}{2 r_1}$

Answer

$\frac{\mu_0 \pi r_1^2}{2 r_2}$

Answer

$\frac{\mu_0 \pi r_2^2}{2 r_1}$

Electromagnetic Induction

The planar concentric rings of metal wire having radii r_1 and r_2 (with r_1>r_2 ) are placed in air. The current I is flowing through the coil of larger radius. The mutual inductance between the coils is given by ( \mu_0= permeability of free space)

Choose the correct answer: