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A 50 turn circular coil of radius 2.0 cm...

A 50 turn circular coil of radius 2.0 cm carrying a current of 5.0 A is rotated in a magnetic field of strength 0.20 T. (a) What is the maximum torque that acts on the coil? (b) in a particular position of the coil, the torque acting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil?

A

`60^(@)`

B

`30^(@)`

C

`90^(@)`

D

`100^(@)`

Text Solution

Verified by Experts

The correct Answer is:
A

`tau = MB sin theta = tau_(max) sin theta`
`(tau_(max))/(2) = tau_(max) sin theta rArr sin theta = (1)/(2)`
`theta = 30^(@)`
Angle between the magnetic field and the plane of the coil
`= 90^(@) - theta = 90^(@) - 30^(@) = 60^(@)`
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