Question 160: Electromagnetic Induction

Question

Two concentric circular coils with radii $$1 \mathrm{~cm}$$ and $$1000 \mathrm{~cm}$$, and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be ___________ $$	imes 10^{-8} \mathrm{H}$$. (Take, $$\pi^{2}=10$$ )

Accepted Answer

The magnetic field $$B_2$$ due to the current $$I_2$$ in the larger coil with 200 turns is given by:
$$B_2 = \frac{N_2 \mu_0 I_2}{2r_2} = \frac{200 \mu_0 I_2}{2 	imes 10}$$
The magnetic flux $$\phi_{1,2}$$ through the smaller coil due to this magnetic field is given by:
$$\phi_{1,2} = N_1 \vec{B}_2 \cdot \vec{A}_1 = N_1 N_2 \frac{\mu_0 I_2}{2 r_2} \cdot \pi r_1^2$$
Since $$\phi_{1,2} = MI_2$$, we can solve for the mutual inductance $$M$$:
$$M = \frac{N_1 N_2 \frac{\mu_0 I_2}{2 r_2} \cdot \pi r_1^2}{I_2}$$
Substituting the given values for $$r_1$$, $$N_1$$, $$r_2$$, and $$N_2$$:
$$M = \frac{10 	imes 200 	imes 4 \pi 	imes 10^{-7} 	imes \pi 	imes (0.01)^2}{2 	imes 10}$$
Simplifying the expression, we get:
$$M = 4 	imes 10^{-8} \mathrm{H}$$
So, the mutual inductance between the two concentric coils is $$4 	imes 10^{-8} \mathrm{H}$$.

Electromagnetic Induction

Two concentric circular coils with radii $1 \mathrm{~cm} and 1000 \mathrm{~cm}, and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be ___________ \times 10^{-8} \mathrm{H}. (Take, \pi^{2}=10$ )

Integer / Numeric Answer Type