1.

The magnetic dipole moment of steel wire of length L is m. It is bent from the middle and arranged as 60°. So the new magnetic dipole moment will be ......

Answer»

`(m)/(sqrt2)`
`(m)/(2)`
`m`
`2m`

Solution :When it is bent magnetic MOMENT of each part is `(m)/(2)`

ANGLE between both is 120°.
`THEREFORE` Resultant mangetic dipole moment,
`= SQRT(( (m)/(2))^(2) = ((m)/(2))^(2) + 2 ((m)/(2) ) ((m)/(2)) xx COS 120^(@) )`
`= sqrt((m^2 )/(4) + (m^2)/( 4) + 2((m^2)/( 4)) (-(1)/(2) ) ) =sqrt((m^2)/(4) ) = (m)/(2)`


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