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What is toroid ? Obtain formula for the magnitude of magnetic field due to current carrying toroid. |
Answer» Solution :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. |
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