Question

In the figure shown ABCDEFGH is a square conducting frame of side $2m$ and resistance $1\Omega /m.$ 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 DH arm is zero
• B. power consumed as heat in the circuit $200watt$
• C. heat produced in DH and BF arm is zero
• D. current in AG and CB arm is equal and its magnitude is $5A$

Answer 1

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

The correct options are
A current in DH arm is zero
B power consumed as heat in the circuit $200watt$
C heat produced in DH and BF arm is zero
D current in AG and CB arm is equal and its magnitude is $5A$

The induced emf in loop ABHFG
$=\frac{d}{dt}\left(BA\right)=A\frac{dB}{dt}=2\times 10=20V$
The induced emf in loop BCDH and DEFH is: $1\times 10=10volt$

KVL in top left loop is ,
$10-(y-z)+(x-y)-2y=0$
$\Rightarrow x-4y+z=-10$............................. (1)
KVL in right loop : $10-2x-(x-y)-(x-z)=0$
$\Rightarrow -4x+y+z=-10$................................. (2)
By equation (1) & (2) it is seen that $x-y=0$
$\Rightarrow$ no current in DH
{This can also be seen by symmetry}
Now, assume $V_B=0$ , & $V_F=V,$
Then $\frac{V+20}{4}+\frac{V-20}{4}+\frac{V-0}{2}=0$
$\Rightarrow V=0\Rightarrow$ no current in FB, and current in AGFEDCBA is $5A$.
$\therefore$ rate of heat production $=\frac{(40{)}^2}{8}=200watt$