1.

A bar magnet has a magnetic moment of 200 A m^(2).The magnet is suspended in a magnetic field of 0.30 N A^(-1)m^(-1) The torque required to rotate the magnet from its equilibrium position through an angle of 30^(@), will be

Answer»

30 N m
`30 sqrt3 N m`
60 N m
`60sqrt3 NM`

Solution :Torque experienced by a magnet suspended in a UNIFORM magnetic FIELD B is given by
`|vectau| = MB sintheta`
Here, `M = 200 Am^(2), B = 0.30 N A^(-1) m^(-1)`
and `THETA = 30^(@)`
`therefore |vectau|200xx0.30xxsin30^(@) implies vectau= 30Nm`


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