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(a) How is a toroid different from a solenoid ? (b) Use Ampere's circuital law to obtain the magnetic field inside a toroid. (c) Show that in an ideal toroid. The magnetic field (i) inside the toroid and (ii) outside the toroidat any point in the open space is zero. |
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Answer» Solution :(a) A toroid can be viewed as a solenoid which has been bent into a circuit shape to close on itself. (b)   For the magnetic FIELD at a point S inside a toroid we have `B( 2 PI r) = mu_(0) N I` `B= mu_(0) (N I)/(2 pi r)= mu_(0)n I` ( `n=` no. of turns per unit length of solenoid ) (c) For the loop I, Ampere.s circuital law gives `B_(1).2pi r_(1) = mu_(0)(0)` i.e., `B_(1)=0` Thus the magnetic field, in the open space inside the toroid is zero. ALSO at point Q, we have `B_(3) ( 2 pi r_(3)) = mu_(0) ( I_("enclosed"))` Butfrom THESECTION cut, we see that the current coming out of the plane of the paper, is cancelled exactly by the current going into it. Hence, `I_("enclosed")=0` `:. B_(3)=0` |
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