Q151 / 160 · AP EAMCET 2024 (21 May Shift 2) · Physics
At a place the horizontal component of earth's magnetic field is $3 \times 10^{-5} \mathrm{~T}$ and the magnetic declination is $30^{\circ}$. A compass needle of magnetic moment $18 \mathrm{Am}^2$ pointing towards geographic north at this place experiences a torque of
- A$36 \times 10^{-5} \mathrm{Nm}$
- B$18 \times 10^{-5} \mathrm{Nm}$
- C$54 \times 10^{-5} \mathrm{Nm}$
- Correct answerD$27 \times 10^{-5} \mathrm{Nm}$
Solution
$\mathrm{B}_{\mathrm{H}}=3 \times 10^{-5} \mathrm{~T}, \phi=30^{\circ}, \mathrm{M}=18 \mathrm{Am}^2$
Torque, $\tau=\mathrm{MB}_{\mathrm{H}} \sin \phi=18 \times 3 \times 10^{-5} \sin 30^{\circ}$ $=27 \times 10^{-5} \mathrm{Nm}$
Torque, $\tau=\mathrm{MB}_{\mathrm{H}} \sin \phi=18 \times 3 \times 10^{-5} \sin 30^{\circ}$ $=27 \times 10^{-5} \mathrm{Nm}$
Hence, the correct answer is option: 'D'.