Torque needed to deflect a magnet through \( 45^\circ \) in the earth's horizontal magnetic field of induction \( 0.5\times 10^{-4}\,\mathrm{T} \) is
(Given the magnetic moment of magnet is \( 4\,\mathrm{A\,m^2} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.41 \times 10^{-4}\,\mathrm{N\,m} \)
A magnet of magnetic moment \( m \) in a uniform field \( B \) experiences a torque that depends on the angle \( \theta \) between \( \vec{m} \) and \( \vec{B} \). To hold it deflected, an equal torque is needed.
\( \displaystyle \tau = m B \sin\theta \)
\( m = 4\,\mathrm{A\,m^2} \)
\( B = 0.5\times10^{-4}\,\mathrm{T} \)
\( \theta = 45^\circ \)
\( \displaystyle \tau = 4 \times 0.5\times10^{-4} \times \sin 45^\circ = 2\times10^{-4} \times 0.707 = 1.41\times10^{-4}\,\mathrm{N\,m} \)