The maximum torque acting on a coil having an effective area \( 0.04\,\mathrm{m^2} \) and carrying current of \( 100\,\mu\mathrm{A} \) is \( 4\times10^{-8}\,\mathrm{N\,m} \), when placed in a magnetic field. The magnitude of the magnetic field is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 10^{-2}\,\mathrm{Wb\,m^{-2}} \)
The torque on a current loop in a uniform field is \( \tau = N I A B\sin\theta \). It is maximum when the plane of the coil is parallel to the field, \( \sin\theta = 1 \). The coil here has an effective area \( A = 0.04\,\mathrm{m^{2}} \) (the turns are included in it).
\( \displaystyle \tau_{\max} = I A B \;\Rightarrow\; B = \frac{\tau_{\max}}{I A} \)
\( I = 100\,\mu\mathrm{A} = 10^{-4}\,\mathrm{A} \)
\( \displaystyle B = \frac{4\times10^{-8}}{10^{-4} \times 0.04} = \frac{4\times10^{-8}}{4\times10^{-6}} = 10^{-2}\,\mathrm{Wb\,m^{-2}} \)