MHT CET · Physics · Electromagnetic Induction
A toroid is a long coil of wire ( \(N\) turns) wound over a circular core. The coefficient of self-induction of the toroid is [The magnetic field in it is uniform and \(R>>r\), where \(r=\) radius of wire, \(R=\) radius of coil] ( \(\mu_0=\) permeability of free space \()\)
- A \(\frac{\mu_0 N^2 R^2}{2 r}\)
- B \(\frac{\mu_0 N r}{2 R}\)
- C \(\frac{\mu_0 N^2 r^2}{R}\)
- D \(\frac{\mu_0 N^2 r^2}{2 R}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mu_0 N^2 r^2}{2 R}\)
Step-by-step Solution
Detailed explanation
We know relation between the flux and the induction is: \(L=\frac{\phi}{I}\)
Flux through the given area \(A\) with magnetic field \(B\) is given by,
\(\phi=N A B\)
The field at the center of the toroid is given by \(B=\mu_0 n I\), where
\(\begin{aligned} & n=\frac{N}{2 \pi R} \\ & \therefore \phi=N\left(\pi r^2\right)\left[\mu_0\left(\frac{N}{2 \pi R}\right) I\right] \\ & \Rightarrow \phi=\frac{\mu_0 N^2 r^2 I}{2 R}\end{aligned}\)
Using the definition of self-induction: \(L=\frac{\phi}{I}=\frac{\mu_0 N^2 r^2}{2 R}\)
Flux through the given area \(A\) with magnetic field \(B\) is given by,
\(\phi=N A B\)
The field at the center of the toroid is given by \(B=\mu_0 n I\), where
\(\begin{aligned} & n=\frac{N}{2 \pi R} \\ & \therefore \phi=N\left(\pi r^2\right)\left[\mu_0\left(\frac{N}{2 \pi R}\right) I\right] \\ & \Rightarrow \phi=\frac{\mu_0 N^2 r^2 I}{2 R}\end{aligned}\)
Using the definition of self-induction: \(L=\frac{\phi}{I}=\frac{\mu_0 N^2 r^2}{2 R}\)
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