Question

A triangular wire frame (each side $=2m$ ) is placed in region of time variant magnetic field having $dB/dt=\sqrt{3}T/s$. The magnetic field is perpendicular to the plane of the triangle. The base of the triangle $AB$ has a resistance $1\Omega$ while the other two sides have resistance $2\Omega$ each. The magnitude of potential difference between the points $A$ and $B$ will be $\frac{a}{10}$ volt then the value of $a$ is.

Answer 1

Answer: 4.00

induced emf in triangle $ABC=\frac{d\varphi }{dt}$
$\varphi =BA$
$e=A\frac{dB}{dt}=\frac{1}{2}\times 2\times \sin 60^{\circ }\times \sqrt{3}$
$=3$ volt
$i=\frac{3}{2+2+1}=\frac{3}{5}$
then is a induced emf in $AB$.

$\underset{A}{\overset{B}{\int }}\vec{E}\cdot d\vec{\ell }=\frac{-d(B\times A\cos \pi )}{dt}$
$e_B-e_A=\frac{dB}{dt}\times A$
$=\sqrt{3}\times \frac{1}{2}\times 2\times \frac{1}{\sqrt{3}}=1$ volt

$V_A-V_B=\frac{3}{5}\times 1-1$
$=\frac{3-5}{5}=\frac{-2}{5}$
$V_B-V_A=0.4$ Volt.