AP EAMCET · PHYSICS · Magnetic Properties of Matter
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 \(54 \times 10^{-5} \mathrm{Nm}\)
- D \(27 \times 10^{-5} \mathrm{Nm}\)
Answer & Solution
Correct Answer
(D) \(27 \times 10^{-5} \mathrm{Nm}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{B}_{\mathrm{H}}=3 \times 10^{-5} \mathrm{~T}, \phi=30^{\circ}, \mathrm{M}=18 \mathrm{Am}^2\) Torque, \(\tau=\mathrm{MB}_{\mathrm{H}} \sin \phi=18 \times 3 \times 10^{-5} \sin 30^{\circ}\) \(=27 \times 10^{-5} \mathrm{Nm}\)
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