Question 25: Electromagnetic Induction

Question

A coil of wire of a certain radius has 100 turns
and a self inductance of 15 mH. The self
inductance of a second similar coil of 500 turns
will be

Accepted Answer

Given, number of turns in the first coil, $$N_1=100$$
Its self-inductance, $$L_1=15 \mathrm{mH}=1.5 	imes 10^{-2} \mathrm{H}$$
Number of turns in the second coil,
$$\begin{array} & N_2 & =500 \
L_2 & =	ext { ? }\end{array}$$
We know that, self inductance of the coil is given by
$$L=\frac{\mu_0\pi N^2R}{2}$$
$$\Rightarrow  L  \propto N^2$$
$$	herefore  \frac{L_2}{L_1}  =\left(\frac{N_2}{N_1}ight)^2=\left(\frac{500}{100}ight)^2=25$$
$$\Rightarrow  L_2 =25 L_1$$
$$=25 	imes 1.5 	imes 10^{-2}$$
$$=37.5 	imes 10^{-2} \mathrm{H}$$
$$=375 	imes 10^{-3} \mathrm{H}=375 ~\mathrm{mH}$$

Answer

75 mH

Answer

375 mH

Answer

15 mH

Answer

None of the above

Electromagnetic Induction

A coil of wire of a certain radius has 100 turns and a self inductance of 15 mH. The self inductance of a second similar coil of 500 turns will be

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