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A circular coil of wire 8 cm in diameter...

A circular coil of wire 8 cm in diameter has 12 turns and carries a current of 5 A. The coil is in a field where the magnetic induction is 0.6 T.
a. What is the maximum torque on the coil?
b. In what position would the torque be half as great as in (i) ?

Text Solution

Verified by Experts

(a) The dipole moment of the coil: `M=NiA`
Here `N=12turns, i=5A and A=pir^2=pi(4xx10^-2)^2`
`:. M=12xx5xxpixx(4xx10^-2)^2=0.3Am^2`
Maximum torque =`MB=0.3xx0.60`
`=0.18Nm`
(b) If `theta` be the axis of the coil and the field
torque=`MB sin theta`
According to the problem, torque=`MB//2`
So, `(MB)/2=MBsin theta implies sin theta=1/2or theta=30^@`
Thus, the normal to the coil is at `30^@` to the field.
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