MP Board · Class 12 · Physics · Electromagnetic InductionA rectangular coil consisting of $N = 200$ turns, area $A = 0.05\text{ m}^2$, and total resistance $R = 5\ \Omega$ is rotated continuously at a speed of $600\text{ rpm}$ about an axis in its plane, perpendicular to a uniform horizontal magnetic field $B = 0.4\text{ T}$. \nCalculate: The angular velocity ($\omega$) of the coil. The maximum magnetic flux ($\Phi{\text{max}}$) linked with the coil. The maximum peak induced electromotive force ($e0$) generated in the coil. The maximum induced current ($I0$) flowing through the coil. The average power loss ($P{\text{avg}}$) due to Joule heating in the coil.

Given Data:

• Number of turns ($N$) = $200$
• Area of coil ($A$) = $0.05\text{ m}^2$
• Resistance ($R$) = $5\ \Omega$
• Rotational speed ($n$) = $600\text{ rpm} = \frac{600}{60}\text{ rps} = 10\text{ rev/s}$
• Magnetic field ($B$) = $0.4\text{ T}$

Step-by-Step Solution:

1. Angular Velocity ($\omega$)

$$\omega = 2 \pi n = 2 \times \pi \times 10 = 20\pi\text{ rad/s} \approx 62.83\text{ rad/s}$$

2. Maximum Magnetic Flux ($\Phi_{\text{max}}$)\nMaximum flux occurs when the plane of the coil is perpendicular to the magnetic field ($\theta = 0^\circ$):

$$\Phi_{\text{max}} = B \times A = 0.4 \times 0.05 = 0.02\text{ Wb}$$

3. Peak Induced EMF ($e_0$)\nAccording to the principle of an AC generator, induced emf is given by $e = e_0 \sin(\omega t)$, where:

$$e_0 = N B A \omega$$ $$e_0 = 200 \times 0.4 \times 0.05 \times (20\pi)$$ $$e_0 = 200 \times 0.02 \times 20\pi = 80\pi\text{ V}$$ $$e_0 \approx 80 \times 3.1416 = 251.33\text{ V}$$

4. Maximum Induced Current ($I_0$)

$$I_0 = \frac{e_0}{R} = \frac{80\pi}{5} = 16\pi\text{ A} \approx 50.27\text{ A}$$

5. Average Power Dissipated ($P_{\text{avg}}$)\nFor an alternating emf wave, root-mean-square voltage is $e_{\text{rms}} = \frac{e_0}{\sqrt{2}}$.

$$P_{\text{avg}} = \frac{e_{\text{rms}}^2}{R} = \frac{(e_0 / \sqrt{2})^2}{R} = \frac{e_0^2}{2R}$$ $$P_{\text{avg}} = \frac{(80\pi)^2}{2 \times 5} = \frac{6400 \pi^2}{10} = 640 \pi^2\text{ W}$$\nUsing $\pi^2 \approx 9.8696$: $$P_{\text{avg}} = 640 \times 9.8696 = 6316.54\text{ W} \approx 6.32\text{ kW}$$

Final Answer:

1. Angular Velocity: $20\pi\text{ rad/s} \approx 62.83\text{ rad/s}$
2. Maximum Flux: $0.02\text{ Wb}$
3. Peak EMF: $80\pi\text{ V} \approx 251.33\text{ V}$
4. Maximum Current: $16\pi\text{ A} \approx 50.27\text{ A}$
5. Average Power Loss: $640\pi^2\text{ W} \approx 6.32\text{ kW}$