What is toroid ? Obtain formula for the magnitude of magnetic field due to current carrying toroid.

Explanation

1. "Toroid is a device having very close wire of insulated wire on a hollow ring".

2. A solenoid bent into a form of a closed ring is called a toroidal solenoid.
3. It carrying a current I.
4. We shall see that the magnetic field in the open space inside (Point p) and exterior to the toroid (Point Q) is zero. The field `vecB` inside the toroid is constant in magnitude for the ideal toroid of closely wound turns.

5. Figure shows a sectional view of the toroid. The direction of the magnetic field inside a clockwise as per the right hand thumb rule for circular loops.
6. Three circular loops 1, 2 and 3 are shown by a dashed lines.
7. By symmetry, the magnetic field should be tangential to each of them and constant in magnitude for a given loop.
8. The circular areas bounded by loops 2 and 3 both cut the toroid, so that each turn off current carrying wire is cut once by the loop 2 and twice by the loop 3.
9. Let the magnetic field along loop 1 be `B_(1)` in magnitude, then in Ampere.s circuital law,
`L=2pir_(1)`
However, the loop encloses no current, so `I_(e)=0`.
Thus,
`thereforeB_(1)(2pir_(1))=mu_(0)I_(e)`
`thereforeB_(1)=(mu_(0)I_(e))/(2pir_(1))`
Substitute `I_(e)=0` in this equation,
`thereforeB_(1)=0`
Thus, the magnetic field at any point P in the open space inside the toroid is zero.
10. Let the magnetic field along loop 3 be `B_(3)` again from Ampere.s law, `L=2pir_(3)`.
11. However from the sectional cut, we see that the current coming out of the plane of the paper is cancelled exactly by the current going into it. Thus, `I_(e)=0`
`thereforeB_(3)(2pir_(3))=mu_(0)I_(e),I_(e)=0`
`thereforeB_(3)=(mu_(0)I_(e))/(2pir_(3)),I_(e)=0`
`thereforeB_(3)=0`
12. Let `L=2pir` at point S and `I_(e)=NI`
`thereforeB(2pir)=mu_(0)I_(e)`
`thereforeB=(mu_(0)(NI))/(2pir)`
Here, `N=2pirn`
where n = number of turn per unit length
`B=(mu_(0)xx2pirnxxI)/(2pir)`
`thereforeB=mu_(0)nI` which is magnetic field inside solenoid.
13. In an ideal toroid the coils are circular.
14. In reality the turns of the toroidal coil form a helix and there is always a small magnetic field external to the toroid.