A magnet of magnetic moment `M` is situated with its axis along the direction of a magnetic field of strength `B`. The work done in rotating it by an angle of `180^(@)` will be
A
`-MB`
B
`+MB`
C
0
D
`+2MB`
Text Solution
Verified by Experts
The correct Answer is:
D
Topper's Solved these Questions
MAGNETISM
ERRORLESS |Exercise Magnet and it s Properties|15 Videos
MAGNETISM
ERRORLESS |Exercise Earth Magnetism|8 Videos
MAGNETIC EFFECT OF CURRENT
ERRORLESS |Exercise Exercise|394 Videos
RAY OPTICS
ERRORLESS |Exercise SET|25 Videos
Similar Questions
Explore conceptually related problems
A magnet of magnetic moment 20 A - m^(2) is freely suspended is a uniform magnetic field of strength 6 T . Work done to rotate it by 60^(@) will be .
A magnet of magnetic moment M is rotated through 360^(@) in a magnetic field H, the work done will be
A "bar" magnet of magnetic moment 1.5J//T is aligned with the direction of a uniform magnetic field of 0.22 T . The work done in turning the magnet so as to align its magnetic moment opposite to the field and the torque acting on it in this position are respectively.
If a magnetic dipole of moment M situated in the direction of a magnetic field B is rotated by 180^(@) , then the amount of work done is
As loop of magnetic moment M is placed in the orientation of unstable equilbirum position in a uniform magnetic field B . The external work done in rotating it through an angle theta is
A magnet of magnetic moment 2JT^(-1) is aligned in the direction of magnetic field of 0.1T . What is the net work done to bring the magnet normal to the magnrtic field?
A bar magnet of magnetic moment M is aligned parallel to the direction of a uniform magnetic field B. Calculate work done to align the magnetic moment (i) opposite to the field (ii) normal to field direction?
