Question

Inside a very long solenoid of radius R, consider the triangular circuit ABC as shown in the figure. Axis of solenoid is perpendicular to the plane of the paper. If the magnetic field inside the solenoid changes at the rate $\frac{dB}{dt}$, then the magnitude of induced emf (AB = BC and AC = 2R).

• A. in the triangular circuit ABC is $R^2\left|\frac{dB}{dt}\right|$
• B. in the triangular circuit ABC is $\frac{R^2}{2}\left|\frac{dB}{dt}\right|$
• C. between the ends of wire AB is $\frac{R^2}{2}\left|\frac{dB}{dt}\right|$, if AC and BC were removed from the circuit
• D. between the ends of wire BC is $\frac{R^2}{2}\left|\frac{dB}{dt}\right|$, if AC and AB were removed from the circuit

Answer 1

Answer: (A), (C), (D)

The correct options are
A in the triangular circuit ABC is $R^2\left|\frac{dB}{dt}\right|$
C between the ends of wire AB is $\frac{R^2}{2}\left|\frac{dB}{dt}\right|$, if AC and BC were removed from the circuit
D between the ends of wire BC is $\frac{R^2}{2}\left|\frac{dB}{dt}\right|$, if AC and AB were removed from the circuit

According to Faraday's law,
Induced e.m.f = - (rate of change of magnetic flux)
Inside the solenoid, flux $\varphi =B\times \text{Area}$
where B is the magnetic flux density.
Thus, induced e.m.f $=-\frac{d}{dt}(B\times \text{Area})=-\text{Area}\times \frac{dB}{dt}=-R^2\frac{dB}{dt}$
(since area of triangle ABC is $R^2$).

Option A is correct.

(C) and (D) The important point is that a changing magnetic flux induces a circulating electric field (from which we derive the induced e.m.f.). In other words, there are no radial components of electric field (radiating outwards or inwards) so that no e.m.f is actually induced in the length AC of the circuit.

Hence, induced e.m.f across AB = induced e.m.f across BC
$=\frac{-R^2}{2}\frac{dB}{dt}$

Option C and D are correct.