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 ......

Options

A. `(m)/(sqrt2)`
B. `(m)/(2)`
C. `m`
D. `2m`

Answer

B

Explanation

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)`