A bar magnet has a magnetic moment 2.5 J...

A bar magnet has a magnetic moment 2.5 J`T^(-4)`- and is placed in a magnetic field of 0.2 T. The work done in turning the magnet from parallel to antiparallel position relative to field direction is

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The correct Answer is:

Work done is changing the oriention of dipole of moment M is a magnetic field B from position `theta_1 " to " theta_2` is given by
`W = MB(cos theta_1 - cos theta_2)`
Here, `theta_1 = 0^@` and `theta_2 = 180^@`
So, `W = 2 MB= 2 xx 2.5 xx 0.2 = 1 J`

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`0.33J,0.33N-m`

`0.66J,06.66B-m`

`0.33J,0`

`0.66J,0`

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?

`0.1J`

`0.2J`

`1J`

`2J`

If a bar magnet of magnetic moment 'm' is freely suspended in a uniform magnetic field B, the work done in rotating the magnet through an angle thete is

`mB(1- sin theta)`

`mB sin theta`

`mB cos B`

`mB(1-costheta)`