Question

The final torque on a coil having magnetic moment 25 A $\mathrm { m } ^ { 2 }$ in a 5 T uniform external magnetic field, if the coil rotates through an angle 60° under the influence of the magnetic field is

• A. 216.5 N m
• B. 108.25 N m
• C. 102.5 N m
• D. 258.1 N m

Answer 1

Answer: (B)

108.25 N m

Answer 2

$| \vec { \tau } | = | \overrightarrow { \mathrm { m } } \times \overrightarrow { \mathrm { B } } | = \mathrm { mB } \sin \theta$

Here, $\mathrm { m } = 25 \mathrm { Am } ^ { 2 } , \theta = 60 ^ { \circ } ; \mathrm { B } = 5 \mathrm {~T}$

$\therefore \tau = 25 \times 5 \times \sin 60 ^ { \circ }$

Or $\tau = 125 \times \frac { \sqrt { 3 } } { 2 } = 108.25 \mathrm { Nm }$