Question

Apply Ampere’s Circuital Law to find the magnetic field both inside and outside of a toroidal solenoid.

Answer 1

A solenoid is bent in such a way its ends are joined together to form a closed ring shape, is called a toroid. The magnetic field has constant magnitude inside the toroid whereas in the interior region (say, at point P) and exterior region (say, at point Q), the magnetic field is zero.

(a) Open space interior to the toroid: 

Let us calculate the magnetic field B_p at point P.

We construct an Amperian loop 1 of radius r₁ around the point P. For simplicity, we take circular loop so that the length of the loop is its circumference.

L₁ = 2π₁ 

Ampere’s circuital law for the loop 1 is

Scince, the loop 1 encloses no current, Ienclosed = 0

This is possible only if the magnetic field at point P vanishes i.e.

\(\vec B_P\) = 0

(b) Open space exterior to the toroid: 

Let us calculate the magnetic field B_Q at point Q. We construct an Amperian loop 3 of radius r₃ around the point Q. The length of the loop is L₃ = 2πr₃ Ampere’s circuital law for the loop 3 is

Since, in each turn of the toroid loop, current coming out of the plane of paper is cancelled by the current going into the plane of paper.

Thus, I_enclosed = 0

This is possible only if the magnetic field at point Q vanishes i.e.
\(\vec B_Q\)= 0

(c) Inside the toroid:

Let us calculate the magnetic field B at point S by constructing an Amperian loop 2 of radius r around the point S. The length of the loop is L₂ = 2πr₂ Ampere’s circuital law for the loop 2 is

Let I be the current passing through the toroid and N be the number of turns of the toroid,

then I _enclosed = NI

The number of turns per unit length is n = \(\frac{n}{2πr_2}\) then the magnetic field at point S is B_s = μ₀nI