Question

In a very long solenoid of radius $R$, if the magnetic field changes at the rate of $\frac{dB}{dt}$, the induced emf for the triangular circuit ABC as shown in the figure is (AB=BC)

• A. $R^2\left(\frac{dB}{dt}\right)$
• B. $4R^2\left(\frac{dB}{dt}\right)$
• C. $\frac{1}{2}{R}^2\left(\frac{dB}{dt}\right)$
• D. $2R^2\left(\frac{dB}{dt}\right)$

Answer 1

The correct option is A $R^2\left(\frac{dB}{dt}\right)$

Magnitude of flux induced, $\mid E\mid =\mid \frac{d\varphi }{dt}\mid$
Since flux, $\varphi =B.A$
$\Rightarrow \mid \frac{d\varphi }{dt}\mid =\mid \frac{d(B.A)}{dt}\mid$
Since area remains constant
$\Rightarrow \frac{d\varphi }{dt}=A\frac{dB}{dt}$
Area of triangle ABC, $A=\frac{1}{2}2R\times R=R^2$
Therefore $\mid E\mid =\frac{d\varphi }{dt}=R^2\frac{dB}{dt}$
Hence option A is correct.