Q149 / 160 · AP EAMCET 2024 (18 May Shift 1) · Physics
A current carrying coil experiences a torque due to a magnetic field. The value of the torque is $80 \%$ of the maximum possible torque. The angle between the magnetic field and the normal to the plane of the coil is
- A$30^{\circ}$
- B$45^{\circ}$
- C$\tan ^{-1}\left(\frac{3}{4}\right)$
- Correct answerD$\tan ^{-1}\left(\frac{4}{3}\right)$
Solution
the torque experience by a current carrying coil in a magnetic field is
$\mathrm{l}=\mathrm{MB} \sin \theta \Rightarrow \mathrm{t}_{\max }=\mathrm{MB}$
Also, $\mathrm{l}=80 \%$ of $\mathrm{i}_{\max }=\frac{4}{5} \mathrm{t}_{\max }=\frac{4}{5} \mathrm{MB}$
$\therefore \mathrm{MB} \sin \theta=\frac{4}{5} \mathrm{MB} \Rightarrow \sin \theta=\frac{4}{5} \quad \therefore \theta=\tan ^{-1}\left(\frac{4}{3}\right)$
Hence, the correct answer is option: 'D'.