Question

A toroid has a non-ferromagnetic wire 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

• A. \( \frac{\mu_0 NI}{\pi (r_1 + r_2)} \)
• B. \( \frac{\mu_0 NI}{(r_2 - r_1)} \)
• C. \( \frac{\mu_0 NI}{(r_1 + r_2)} \)
• D. \( \frac{\mu_0 NI}{\pi (r_2 - r_1)} \)

Hint:

In toroidal magnetic fields, the field inside the toroid is proportional to the number of turns and inversely proportional to the radius.

Answer 1

The Correct Option is A

Step 1: Understanding the formula for magnetic field inside a toroid.
The magnetic field inside a toroid is given by: \[ B = \frac{\mu_0 NI}{2 \pi r} \] Where \( r \) is the radius of the toroid, and \( N \) is the number of turns. The formula for the field in the center of the toroid with inner and outer radii is: \[ B = \frac{\mu_0 NI}{\pi (r_1 + r_2)} \] Thus, the correct answer is (A) \( \frac{\mu_0 NI}{\pi (r_1 + r_2) \)}.