Home
Class 12
PHYSICS
A bar magnet is hung by a thin cotton th...

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60^(@) is W. Now the torrue required to keep the magnet in this new position is

A

`(3sqrtW)/2`

B

`(2W)/sqrt3`

C

`W/sqrt3`

D

`sqrt3W`

Text Solution

Verified by Experts

The correct Answer is:
D
`tau = PE sin 60^@" "...(i)`
` W = PE (1 - cos 60^@) " ".... (ii)`
From (i) and (ii)
`tau/W=(sqrt3//2)/(1//2)implies tau =Wsqrt3`
Promotional Banner

Similar Questions

Explore conceptually related problems

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in' equilibrium state. The energy required to rotute it by 60^(circ) is W . Now the torque required to keep the magnet in this new: position is sqrt(k) W . Find k .

A bar magnet hung by a insulating thread In a uniform horizontal magnetic field and magnet is in equilibrium Initially. The torque required to keep the magnet at angle by 30° with the direction of magnetic field is t. Then the work done to rotate it by same angle from initial position is

A bar magnet of magnetic moment M is hung by a thin cotton thread in a uniform magnetic field B. Work done by the external agent to rotate the bar magnet from stable equilibrium position to 120° with the direction of magnetic field is (consider change in angular speed is zero)

A magnetic needle lying parallel to a magnetic field requires W units of work to turn through 60^(@) . The external torque required to maintain the magnetic needle in this position is