A magnet is hung horizontally in the magnetic meridian by a wire without any twist. If the supporting wire is given a twist of 180° at the top, the magnet rotates by 30°. Now if another magnet is used, then a twist of 270° at the supporting end of wire also produces a rotation of the magnet by 30°. Compare the magnetic dipole moments of the two magnets.

Options

A. `(4)/(8)`
B. `(8)/(5)`
C. `(5)/(8)`
D. `(8)/(4)`

Answer

C

Explanation

If resultant twist in the wire `= delta`
`delta_(1) = 180^(@) - 30^(@) = 150^(@) xx (pi)/(180) "rad" and `
`delta_(2) = 270^(@) - 30^(@) = 240^(@) =240 xx (pi)/(180) "rad"`
If the twist - constant for the wire is `k` then Rotating torque, `tau_(1) = k delta_(1) and tau_(2) = k delta_(2)`
`therefore (tau_1)/( tau_2) = (delta_1)/( delta_2)`
Here `alpha` is the angle made by the magnetic dipole moment with the magnetic meridian
`tau_(1). =m_(1) B_(H) sin alpha`
Since the second magnet is also rotated by the same angle
`tau_(2). = m_(2) B_H sinalpha`
`therefore (tau_(1).)/(tau_(2) .) = (m_1)/(m_2)`
At equilibrium `tau_(1) = tau_(1) . and tau_(2)= tau_(2).`
`(tau_(1).) /( tau_(2) .) = (tau_(1))/( tau_(2))`
`(m_1)/(m_2) = (delta_1)/( delta_2) = (150)/(240) = (5)/(8)`