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

Answer: (B)

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

$area=\frac{\sqrt{3}{a}^2}{4}$
$e=\frac{-d\varphi }{dt}=B_o\lambda \frac{\sqrt{3}{a}^2}{4}\times e^{-\lambda t}$
$i=\frac{e}{3R}$
$=\frac{B_o\lambda a^2}{4\sqrt{3}R}{e}^{-\lambda t}$