Question

Show that in an ideal toroid the magnetic field outside the toroid at any point in the open space is zero.

Answer 1

Hint A long closely wound helical coil is called a solenoid. A solenoid bend in the form of a closed ring is called a toroid. The magnitude of the magnetic field $B$ will be the same at all points on the circular axis of the toroid. The direction of the magnetic field will be tangential to the circular axis of the toroid.
Formula used:
Ampere’s circuital law
$\oint {\overrightarrow B .\overrightarrow {dl} = {\mu _0}I} $
Where, $I$ stands for the current through the conductor, $dl$ is an elementary length of the loop, $B$ stands for the magnetic induction at the centre of $dl$ and ${\mu _0}$is the permeability of free space.

Complete Step by step solution
Consider an amperian loop along the axis of a toroid. Let us consider $B$ to be the magnetic field along the amperian loop.
By applying the Ampere’s circuital theorem
$\oint {\overrightarrow B .\overrightarrow {dl} = N{\mu _0}I} $
Where $N$ stands for the total number of turns of the toroid.
On integrating the L.H.S of the above equation, we get
$B\oint {\overrightarrow {dl} = B.2\pi r} $ $\because \int {\overrightarrow {dl} = 2\pi r} $ (Where, $r$stands for the radius of the toroid)
$B2\pi r = {\mu _0}NI$
From this we get,
$B = \dfrac{{{\mu _0}NI}}{{2\pi r}}$
This can be written as,
$B = {\mu _0}I\dfrac{N}{{2\pi r}}$
Where $\dfrac{N}{{2\pi r}}$stands for the number of turns per unit length. Let $\dfrac{N}{{2\pi r}} = n$
Then,
$B = {\mu _0}nI$

The number of turns per unit length is zero outside the toroid at any point. So the magnetic field will be zero at points outside and inside the toroid.

Note
Ampere’s circuital law is an alternative way of expressing Biot Savart’s law. Ampere’s circuital law states that the line integral of magnetic field around a closed path in free space is equal to ${\mu _0}$ times the current enclosed by the path. Ampere’s circuital law is analogous to Gauss’s law in electrostatics.