Question

Figure shows a square loop of side 5 cm being moved towards right at a constant speed of 1 cm/s. The front edge enters the 20 cm wide magnetic field at t = 0. Find the total heat produced in the loop during the interval 0 to 30 s if the resistance of the loop is 4.5 mΩ.

Answer 1

Resistance of the coil, R = 45 mΩ = 4.5 × 10⁻³ Ω
The heat produced is found by taking the sum of the individual heats produced.
Thus, the net heat produced is given by
H = H₁ + H₂ + H₃ + H₄

(a) Heat developed for the first 5 seconds:-
 Emf induced, e = 3 × 10⁻⁴ V
Current in the coil,

\[i = \frac{e}{R} = \frac{3 \times {10}^{- 4}}{4 . 5 \times {10}^{- 3}}\]= 6.7 × 10⁻² A

H₁ = (6.7 × 10⁻²)² × 4.5 × 10⁻³ × 5
There is no change in the emf from 5 s to 20 s and from 25 s to 30 s.
Thus, the heat developed for the above mentioned intervals is given by
H₂ = H₄ = 0

Heat developed in interval t = 25 s to 30 s:-
The current and voltage induced in the coil will be the same as that for the first 5 seconds.
H₃ = (6.7 × 10⁻²)² × 4.5 × 10⁻³ × 5
Total heat produced:-
H = H₁ + H₃
    = 2 × (6.7 × 10⁻²)² × 4.5 × 10⁻² × 5
    = 2 × 10⁻⁴ J