Question

, `ABCEFGA` is a square conducting frame of side `2 m` 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 `10 T s^(-1)`. Find the rate at which heat is produced in the circuit, `AB = BC = CD = BH`.

Answer 1

Correct Answer - `200W`
Induced emfs are `10 V`, `10 V and 20 V` as shown.

Top left loop: `10 = (I_(1) - I_(2)) xx 1 + (I - I_(2)) xx 2 - I_(2)`
`rarr 10 = 2I - 4I_(2) + I_(1)` (i)
Top right loop: `10 = (I - I_(1)) xx 1 + (I - I_(1)) xx 1 - (I_(1) - I_(2)) xx 1 - I_(1) xx 1`
`rarr 10 = 2I - 4I_(1) + I_(2)` (ii)
from (i) and (ii) `I_(1) - I-(2)`
Bottom loop: `20 = 2I + 1I + 1I_(1) + 1I_(2) xx 1I`
`rarr 20 = 4I + I_(1) + I_(2)`
`rarr 20 = 4I + 2I_(1)` (iii)
From (ii) `10 = 2I - 3I_(1)` rarr `20 = 4I - 6I_(1)` (iv)
From (iii) and (iv), `I_(1) = 0`, So `I_(2) = 0`, then from (iii) `I = 5 A`
So a current of `5 A` flows in outer branch and no current in the inner branches.
`"Heat produced" = I^(2)R_(outer branches) = 5^(2) xx 8 = 200 W`