First slide

Magnetic dipole in a uniform magnetic field

Question

A bar magnet is held perpendicular to a uniform field. If the couple acting on the magnet is to be halved by rotating it, the angle by which it is to be rotated is

Moderate

A

30^o
B

45^o
C

60^o
D

90^o

Solution

Given that $θ_{1} = 90^{\circ}$
$∴$ Initial torque $τ_{1} = τ$
Final torque $τ_{2} = τ / 2$
$Now τ = τ_{1} = MBsin θ_{1} = MBsin 90^{\circ} = MB \begin{aligned} and τ_{2} = MBsin θ_{2} \\ or τ / 2 = MBsin θ_{2} \end{aligned} ∴ \frac{MB}{2} = MBsin θ_{2} or \sin θ_{2} = \frac{1}{2} ∴ θ_{2} = 30^{\circ}$
So, the angle by which the bar magnet is to be rotated $(θ_{2}) = 30^{o}$ . $d θ = θ_{1} - θ_{2} = 90^{0} - 30^{0} = 60^{0}$

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