For a toroid N=500 , radius =40 ~cm , and area of cross section =10 ~cm^2 . Find inductance.

Correct answer: A. 125 μ H If we flow a current i , then aligned & φ_ self =(B × A) × N=((μ_0 N i)/(2 π R)) A × N & φ_ self =((μ_0 N^2 A)/(2 π R)) i and φ_ self =L i aligned aligned & L=(μ_0 N^2 A)/(2 π R)=(μ_0)/(4 π) (2 N^2 A)/(R) & L=(10^ -7 ) × 2 ×(500)^2 × 10 ×(10^ -2 )^2 0.4 &=125 μ H aligned

AIIMS 201926 May

Exam: AIIMS
Year: 2019
Date: 26 May
Subject: Physics
Chapter: Electromagnetic Induction

AIIMS 2019 · 26 May · Physics · Electromagnetic Induction

For a toroid N=500 , radius =40 ~cm , and area of cross section =10...

For a toroid

N=500

, radius

=40 \mathrm{~cm}

, and area of cross section

=10 \mathrm{~cm}^2

. Find inductance.

Options

A $125 \mu \mathrm{H}$
B $250 \mu \mathrm{H}$
C $0.00248 \mathrm{H}$
Dzero

Correct answer

A. $125 \mu \mathrm{H}$

Step-by-step solution

If we flow a current

i

, then

\begin{aligned} & \phi_{\text{self }}=(B \times A) \times N=\left(\frac{\mu_0 N i}{2 \pi R}\right) A \times N \\ & \Rightarrow \quad \phi_{\text{self }}=\left(\frac{\mu_0 N^2 A}{2 \pi R}\right) i \text{\text{ and } } \phi_{\text{self }}=L i\end{aligned}

\begin{aligned} & L=\frac{\mu_0 N^2 A}{2 \pi R}=\frac{\mu_0}{4 \pi} \frac{2 N^2 A}{R} \\ & \Rightarrow L=\left(10^{-7}\right) \times \frac{2 \times(500)^2 \times 10 \times\left(10^{-2}\right)^2}{0.4} \\ &=125 \mu \mathrm{H}\end{aligned}

Practice Electromagnetic Induction on Quantrex Academy →

More from Electromagnetic Induction