A bar magnet of magnetic moment 220 Am 2 is suspended in a magnetic field of intensity 0.25 N / Am. The couple required to deflect it through 30° is:

Question

A bar magnet of magnetic moment 220 Am 2 is suspended in a magnetic field of intensity 0.25 N / Am. The couple required to deflect it through 30° is: (A) 27.5 Nm (B) 20.25 Nm (C) 47.63 Nm (D) 12 Nm

Tardigrade Admin · Accepted Answer

A bar magnet suspended in a uniform magnetic field sets itself with its axis parallel to the field. A magnet placed in the magnetic field experiences a torque which rotates the magnet to a position in which the axis of the magnet is parallel to the field.
The magnitude of torque acting on a current loop placed in a magnetic field

B

with its axis at an angle

θ

with the direction of

B

is given by

τ = i A B sin θ

Here, magnitude of dipole moment,

M = i A

∴ τ = MB sin θ

Putting the numerical values, we have

M = 220 A m^{2}, B = 0.25 N / A m, θ = 3 0^{\circ}

∴ τ = 220 \times 0.25 \times cos 3 0^{\circ}

= 220 \times 0.25 \times \frac{3}{2}

= 47.63 N m

Q. A bar magnet of magnetic moment 220Am2 is suspended in a magnetic field of intensity 0.25N/Am. The couple required to deflect it through 30∘ is:

27.5 Nm

20.25 Nm

47.63 Nm

12 Nm

Q. A bar magnet of magnetic moment $220 A m^{2}$ is suspended in a magnetic field of intensity $0.25 N / A m$ . The couple required to deflect it through $3 0^{\circ}$ is: