Question

In the figure shown $ABCDEFGA$ is a square conducting frame of side $2m$ and resistance $1\Omega /m$. The conducting wires $BFandDH$ of the same material are also connected to the frame as shown in the figure. A uniform magnetic field $B$ is applied perpendicular to the plane and pointing inwards. It increases with time at a constant rate of $10T/s$. Then ($AB=BC=CD=BH=1m$)

• A. Current in $BH$ arm is zero
• B. Power dissipated as heat in the circuit $200$ watt.
• C. Heat produce in $DH\&BF$ arm is zero
• D. Current in $AG\&CB$ arm is equal its magnitude is $5A$.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Power dissipated as heat in the circuit $200$ watt.
B Current in $BH$ arm is zero
C Heat produce in $DH\&BF$ arm is zero
D Current in $AG\&CB$ arm is equal its magnitude is $5A$.

The induced emf in loop $ABHFG$

$=\frac{d}{dt}\left(\text{BA}\right)=\text{A}\frac{d\text{B}}{dt}=2\times 10=20V$

The induced emf in loop $BCDH\&DEFH$

$=1\times 10=10\text{volt}$.

$KVL$ in top left loop is,

$=1\times 10=10\text{volt}$.

$10-\text{(y - z)+(x -y)}-2\text{y}=0$

$\Rightarrow \text{x}-4\text{y}+\text{z}=-10......\left(1\right)$

$KVL$ in right loop : $10-2\text{x}-\text{(x - y) - (x -z)}=0$

$\Rightarrow -4\text{x}+\text{y}+\text{z}=-10......\left(2\right)$

By equation (1) & (2) it is seen that $\text{x - y}=0\Rightarrow \text{no current in DH}$

$\left\{\begin{array}{c}\text{This can also be seen by symmetry}\end{array}\right\}$

This makes solution very simple now the circuit is,

Assume ${\text{v}}_{\text{B}}=0,\&{\text{v}}_{\text{F}}=\text{v}$,

Then $\frac{\text{v}+20}{4}+\frac{\text{v}-20}{4}+\frac{\text{v}-0}{2}=0$

$\Rightarrow \text{v}=0\Rightarrow \text{no current in FB, and current in AGFEDCBA is 5 A}$

$\therefore \text{Circuit is}$

$\therefore \text{Rate of heat production}=(40{)}^2/8=200\text{watt}$.