Question 227: Electromagnetic Induction

Question

Two planar concentric rings of metal wire having radii $\mathrm{r}_1$ and $\mathrm{r}_2\left(\mathrm{r}_1>\mathrm{r}_2\right)$ 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
$$ B=\frac{\mu_0 N I}{2 R} $$
$	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 $\mathrm{B}_{\mathrm{A}}$,
$$ \begin{aligned} & \phi_B=B_A A_B \ & A_B=\pi r_2^2 \end{aligned} $$
$$ 	herefore \quad \phi_B=\frac{\mu_0 I}{2 r_1} 	imes \pi r_2^2=\frac{\mu_0 \pi r_2^2}{2 r_1} I $$
Mutual Inductance $M=\frac{\phi}{I}$
$	herefore \quad$ We can write,
$$ M=\frac{\phi_B}{I}=\frac{\mu_0 \pi r_2^2 \cdot I}{2 r_1 \cdot I}=\frac{\mu_0 \pi r_2^2}{2 r_1} $$

Answer

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

Answer

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

Answer

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

Answer

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

Electromagnetic Induction

Two planar concentric rings of metal wire having radii \mathrm{r}_1 and \mathrm{r}_2\left(\mathrm{r}_1>\mathrm{r}_2\right) 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: