Question

A triangular wire frame (each side $=2\, m$ ) 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

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

$\int\limits_{A}^{B} \vec{E} \cdot d \vec{\ell}=\frac{-d(B \times A \cos \pi)}{d t}$
$e_{B}-e_{A} =\frac{d B}{d t} \times A$
$=\sqrt{3} \times \frac{1}{2} \times 2 \times \frac{1}{\sqrt{3}}=1$ 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.