Question

Three resistances of magnitude $R$ each are connected in the form of an equilateral triangle of side $a$. The combination is placed in a magnetic field $B=B_0{e}^{-\lambda t}$ perpendicular to the plane. The induced current in the circuit is given by

• A. $\left(\frac{a^2\lambda }{2\sqrt{3}R}{B}_0\right)e^{-\lambda t}$
• B. $\left(\frac{a^2\lambda }{4\sqrt{3}R}{B}_0\right)e^{-\lambda t}$
• C. $\left(\frac{a^2{B}_0}{4\lambda \sqrt{3}R}\right)e^{-\lambda t}$
• D. $\left(\frac{a^2{B}_{0}R}{4\lambda \sqrt{3}}\right)e^{-\lambda t}$

Answer 1

The correct option is B $\left(\frac{a^2\lambda }{4\sqrt{3}R}{B}_0\right)e^{-\lambda t}$

emf generated is ${ϵ}_{in}=NA\cos \theta \frac{dB}{dt}$
current generated is
$I_{in}=\frac{{ϵ}_{in}}{R_{net}}=\frac{NA\cos 0}{R_{net}}\frac{dB}{dt}$
$I_{in}=\frac{\sqrt{3}}{4}\frac{a^2}{3R}\frac{d}{dt}\left(B_0{e}^{-\lambda t}\right)=\frac{a^2{B}_0\lambda }{4\sqrt{3}R}\left(e^{-\lambda t}\right)$