Question

A small coil is introduced between the poles of an electromagnet so that its axis coincides with the magnetic field direction. The number of turns is \(n\) and the cross-sectional area of the coil is \(A\). When the coil turns through \(180^\circ\) about its diameter, the charge flowing through the coil is \(Q\). The total resistance of the circuit is \(R\). What is the magnitude of the magnetic induction?

• A. \( \dfrac{QR}{nA} \)
• B. \( \dfrac{2QR}{nA} \)
• C. \( \dfrac{Qn}{2RA} \)
• D. \( \dfrac{QR}{2nA} \)

Hint:

When coil is flipped by \(180^\circ\), flux changes from \(BA\) to \(-BA\), so \(\Delta\Phi = 2BA\). Use \(Q=\dfrac{n\Delta\Phi}{R}\).

Answer 1

The Correct Option is D

Step 1: Use Faraday’s law in terms of charge.
Induced emf:
\[ \mathcal{E} = -n\frac{d\Phi}{dt} \] Current:
\[ i = \frac{\mathcal{E}}{R} \] Charge flowed:
\[ Q = \int i\,dt = \frac{n}{R}\int d\Phi = \frac{n}{R}\Delta\Phi \] Step 2: Change in flux when rotated by \(180^\circ\).
Initial flux:
\[ \Phi_i = BA \] Final flux after \(180^\circ\):
\[ \Phi_f = -BA \] So,
\[ \Delta\Phi = \Phi_f - \Phi_i = -BA - BA = -2BA \] Magnitude:
\[ |\Delta\Phi| = 2BA \] Step 3: Substitute in charge formula.
\[ Q = \frac{n}{R}(2BA) \Rightarrow B = \frac{QR}{2nA} \] Final Answer: \[ \boxed{\dfrac{QR}{2nA}} \]