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 60o is W, Now the torque required to keep the magnet in this new position is.
√3W
√32W
W√3
2W√3
A
2W√3
B
W√3
C
√3W
D
√32W
Open in App
Solution
Verified by Toppr
Torque acting on a bar magnet kept in a magnetic field →B is τ=→m×→B=mBsinθ
Hence work done in moving from θ=0∘ to θ=60∘ is W=mBcos60∘=12mB
Hence torque acting when magnet is placed at 60∘=√32mB=√3W
Was this answer helpful?
40
Similar Questions
Q1
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 60o is W, Now the torque required to keep the magnet in this new position is.
View Solution
Q2
The work required to rotate a magnetic needle by 600from equilibrium position in a uniform magnetic field is W. The torque required to hold it in that position is
View Solution
Q3
A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60∘. The torque required to maintain the needle in this position will be
View Solution
Q4
A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of 60∘ and the work is W. The torque on the magnetic needle at this position is:
View Solution
Q5
A magnetic needle lying parallel to a magnetic field requires W unit of work to turn it through 60∘. The torque needed to maintain the needle in this position is