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

Correct answer: B. 375 mH Inductance of a coil is given by aligned & L=(1)/(2) μ_0 π N^2 R (L_2)/(L_1)=(N_2^2)/(N_1^2) & L_2=L_1 (N_2^2)/(N_1^2)=((500)/(100))^2 15 mH=375 mH aligned

AIIMS 2009

Exam: AIIMS
Year: 2009
Subject: Physics
Chapter: Electromagnetic Induction

AIIMS 2009 · Physics · Electromagnetic Induction

A coil of wire of a certain radius has 100 turns and a self...

A coil of wire of a certain radius has 100 turns and a self inductance of

15 \mathrm{mH}

. The self inductance of a second similar coil of 500 turns will be

Options

A $75 \mathrm{mH}$
B $375 \mathrm{mH}$
C $15 \mathrm{mH}$
Dnone of these

Correct answer

B. $375 \mathrm{mH}$

Step-by-step solution

Inductance of a coil is given by

\begin{aligned} & L=\frac{1}{2} \mu_0 \pi N^2 R \Rightarrow \frac{L_2}{L_1}=\frac{N_2^2}{N_1^2} \\ & \therefore L_2=L_1 \frac{N_2^2}{N_1^2}=\left(\frac{500}{100}\right)^2 15 \mathrm{mH}=375 \mathrm{mH}\end{aligned}

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