CBSE · Class 12 · Physics · Electromagnetic InductionA square coil of side $10\text{ cm}$ consists of 500 turns and is placed perpendicular to a uniform magnetic field of $0.4\text{ T}$. If the coil is rotated through $180^\circ$ about an axis perpendicular to the field in $0.1\text{ s}$, calculate the magnitude of the induced electromotive force (emf) in the coil.

Step-by-Step Solution:

1. Given Data:

• Side of the square coil, $a = 10\text{ cm} = 0.1\text{ m}$
• Area of the coil, $A = a^2 = (0.1\text{ m})^2 = 0.01\text{ m}^2$
• Number of turns, $N = 500$
• Magnetic field strength, $B = 0.4\text{ T}$
• Time taken for rotation, $\Delta t = 0.1\text{ s}$
• Initial orientation angle, $\theta_1 = 0^\circ$
• Final orientation angle, $\theta_2 = 180^\circ$

2. Initial and Final Magnetic Flux:

• Initial flux linked with one turn: $$\Phi_1 = B A \cos(0^\circ) = 0.4 \times 0.01 \times 1 = 0.004\text{ Wb}$$
• Final flux linked with one turn: $$\Phi_2 = B A \cos(180^\circ) = 0.4 \times 0.01 \times (-1) = -0.004\text{ Wb}$$

3. Change in Flux:

• Change in flux for 1 turn, $\Delta \Phi = \Phi_2 - \Phi_1 = -0.004 - 0.004 = -0.008\text{ Wb}$
• Total flux change for $N$ turns: $$\Delta \Phi_{\text{total}} = N \Delta \Phi = 500 \times (-0.008\text{ Wb}) = -4.0\text{ Wb}$$

4. Induced EMF Calculation:\nUsing Faraday's Law of Electromagnetic Induction: $$e = -\frac{\Delta \Phi_{\text{total}}}{\Delta t}$$ $$e = -\frac{-4.0\text{ Wb}}{0.1\text{ s}} = 40\text{ V}$$

Answer:\nThe magnitude of the induced emf in the coil is $40\text{ V}$.