Question 294: Electromagnetic Induction

Question

Two concentric circular coils having radii $$r_1$$ and $$r_2\left(r_2 << r_1\right)$$ are placed co-axially with centres coinciding. The mutual induction of the arrangement is (Both coils have single turn, $$\mu_0=$$ permeability of free space)

Accepted Answer

Let $$I_1$$ be the current through the coil whose radius is $$r_1$$.
$$	herefore \quad$$ Magnetic field at the centre of the coil,
$$\mathrm{B}_1=\frac{\mu_0 \mathrm{I}_1}{2 \mathrm{r}_1}$$
Magnetic flux passing through the coil of radius $$r_2$$ is
$$\begin{aligned}
& \phi_2=\mathrm{B}_1 \cdot \pi \mathrm{r}_2^2 \quad (\because \phi=	ext { B.A }) \
& =\frac{\mu_0 \mathrm{I}_1}{2 \mathrm{r}_1} \cdot \pi \mathrm{r}_2^2
\end{aligned}$$
$$	herefore \quad$$ Mutual inductance of the arrangement,
$$\begin{aligned}
M & =\frac{\phi_2}{I_1} \
& =\frac{\mu_0 I_1 \pi r_2^2}{2 r_1 I_1} \
& =\frac{\mu_0 \pi r_2^2}{2 r_1}
\end{aligned}$$

Answer

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

Answer

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

Answer

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

Answer

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

Electromagnetic Induction

Two concentric circular coils having radii $r_1 and r_2\left(r_2 << r_1\right) are placed co-axially with centres coinciding. The mutual induction of the arrangement is (Both coils have single turn, \mu_0=$ permeability of free space)

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