At a place the horizontal component of earth's magnetic field is $3 \times 10^{-5} \mathrm{~T}$ and the…

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
  1. $36 \times 10^{-5} \mathrm{Nm}$
  2. $18 \times 10^{-5} \mathrm{Nm}$
  3. $54 \times 10^{-5} \mathrm{Nm}$
  4. $27 \times 10^{-5} \mathrm{Nm}$

Solution

$\mathrm{B}_{\mathrm{H}}=3 \times 10^{-5} \mathrm{~T}, \phi=30^{\circ}, \mathrm{M}=18 \mathrm{Am}^2$
Torque, $\tau=\mathrm{MB}_{\mathrm{H}} \sin \phi=18 \times 3 \times 10^{-5} \sin 30^{\circ}$ $=27 \times 10^{-5} \mathrm{Nm}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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