MHT CET · Physics · Magnetic Effects of Current
A toroid has a core of inner radius \(r_1\) and outer radius \(r_2\), around which \(N\) turns of wire are wound. If the current in the wire is \(I\) then the magnetic field inside the toroid is ( \(\mu_0=\) permeability of free space)
- A \(\frac{\mu_0 N I}{2 \pi\left(r_1+r_2\right)}\)
- B \(\frac{\mu_0 N I}{\pi\left(r_1+r_2\right)}\)
- C \(\frac{\mu_0 N I}{2 \pi\left(r_2-r_1\right)}\)
- D \(\frac{\mu_0 N I}{\pi\left(r_2-r_1\right)}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mu_0 N I}{\pi\left(r_1+r_2\right)}\)
Step-by-step Solution
Detailed explanation
Considering Ampearian loop at the centre of toroid:
\(\int \overrightarrow{B . d l}=\mu_0 N I\)
Since, \(B\) has radial symmetry \(\&\) angle between \(B \& d l\) is zero
\(\begin{aligned} & \therefore B \int d l=B 2 \pi\left\{\frac{r_1+r_2}{2}\right\}=\mu_0 N I \\ & \Rightarrow B=\left[\frac{\mu_0 N I}{\pi\left(r_1+r_2\right)}\right]\end{aligned}\)
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