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

  $\frac{\mu_0 N I}{2 \pi\left(r_1+r_2\right)}$
• Correct answer
  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

  $\frac{\mu_0 N I}{\pi\left(r_2-r_1\right)}$

Solution

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 $\begin{aligned} & \therefore B \int d l=B 2 \pi\left\{\frac{r_1+r_2}{2}\right\}=\mu_0 N I \\ & \Rightarrow B=\left[\frac{\mu_0 N I}{\pi\left(r_1+r_2\right)}\right]\end{aligned}$

Hence, the correct answer is option: 'B'.