Question

In figure, $ABCEFGA$ is a square conducting frame of side $2m$ and resistance $1\Omega$ $m^{-1}$. A uniform magnetic field $B$ is applied perpendicular to the plane and pointing inward. It increases with time at a constant rate of $10T$ $s^{-1}$. Find the rate at which heat is produced in the circuit, $AB=BC=CD=BH$

Answer 1

emf in the loop BCDH =$m^2\times 10V$. As B increases with time, to counter this effect, by Lenz's law, current in this loop flows in the direction CBHDC(counterclockwise).value of curren in this loop =$i_1=\frac{10m^2}{m^{-1}4m}=\frac{10m^2}{4}A$
emf in the loop DEFH = $m^2\times 10V$. As B increases with time, to counter this effect, by Lenz's law, current in this loop flows in the direction DHFED(counterclockwise). Value of current in this loop = $i_2=\frac{10m^2}{m-14m}=\frac{10m^2}{4}A$
emf in the loop ABFG = $2m^2\times 10V$.As B increases with time, to counter this effect, by Lenz's law current in this loop flows in the direction BAGFB(counterclockwise). Value of current in this loop =$i_3=\frac{20m^2}{m^{-1}6m}=\frac{20m^2}{6}$
Therefore net current in each segment is as in figure.
Heat is produced at the rate = $i_3^2{R}_{AB}+i_1^2{R}_{BC}+i_1^2{R}_{CD}+i_2^2{R}_{DE}+i_2^2{R}_{EF}+i_3^2{R}_{FG}+i_3^2{R}_{AG}+(i_1-i_2{)}^2{R}_{DH}+(i_1-i_3{)}^2{R}_{BH}+(i_2-i_3{)}^2{R}_{FH}=\frac{425m^4}{6}W$