Obtain with the help of a necessary diagram, the expression for the magnetic field in the interior of a toroid carrying current.

To obtain the expression for the magnetic field inside a toroid carrying current, let's consider a toroidal solenoid, which is a hollow circular ring wound uniformly with NN turns of wire carrying a current II.

Let's denote:

• BB as the magnetic field inside the toroid,
• μμ as the permeability of the medium,
• II as the current passing through the toroid,
• NN as the total number of turns of wire in the toroid,
• rr as the radius of the toroid,
• RR as the radius of the cross-section of the toroid.

Now, let's consider a circular path inside the toroid, which is concentric with the toroid itself.

The line integral of BB along this circular path is equal to the permeability of the medium times the total current passing through the surface enclosed by the path.

∮B⋅dl=μIenc∮B⋅dl=μIenc

Where IencIenc is the total current passing through the surface enclosed by the circular path.

Since the current is uniformly distributed along the wire of the toroid, the total current passing through the circular path is equal to the current II multiplied by the total number of turns NN.

Ienc=NIIenc=NI

Therefore, we have:

∮B⋅dl=μNI∮B⋅dl=μNI

By symmetry, the magnetic field BB at every point along the circular path is tangential to the path and has the same magnitude.

The line integral ∮B⋅dl∮B⋅dl along the circular path is equal to BB multiplied by the circumference of the circular path.

B⋅2πr=μNIB⋅2πr=μNI

From this equation, we can solve for BB, the magnetic field inside the toroid:

B=μNI2πrB=2πrμNI

Therefore, the expression for the magnetic field inside the toroid is:

B=μNI2πrB=2πrμNI

This expression shows that the magnetic field inside the toroid is inversely proportional to the radius rr of the circular path and directly proportional to the total current II passing through the toroid and the number of turns NN.