Question

Suppose the ends of the coil in the previous problem are connected to a resistance of 100 Ω. Neglecting the resistance of the coil, find the heat produced in the circuit in one minute.

Answer 1

Use Q = mCΔT

Explanation:

H = $\int_{0}^{t}i^2RdT = \int_{0}^{t}\frac {B^2A^2\omega^2}{R^2} sin \omega tR\ dt$

= $\frac {B^2A^2\omega^2}{2R^2}\int_{0}^{t}(1 - cos2\omega t)dt$

= $\frac {B^2A^2\omega^2}{2R}(t - \frac {sin2\omega t}{2\omega})_0^1^m^i^n^u^t^e$

= $\frac {B^2A^2\omega^2}{2R}(60 - \frac {sin2\omega t}{2\omega})_0^1^m^i^n^u^t^e^$

= $\frac {B^2A^2\omega^2}{2R}(60 - \frac {sin2 \times 8\times \frac {2\pi}{60}\times60}{2\times80\times2\pi/60})$

= $\frac {60}{200} \times \pi^2r^4 \times B^2 \times (80^4\times \frac {2\pi}{60})^2$

= $\frac {60}{200} \times 10 \times \frac {64}{9} \times 10 \times 625 \times 10^-^8 \times 10^-^4$

= $\frac {625\times 6\times 64}{9\times 2}\times 10^-^1^1$

= $1.33 \times 10^-^7$ J