NCERT · Class 12 · Physics · Electromagnetic Induction(a) A square coil of side $10\text{ cm}$ consisting of $500$ turns is placed perpendicular to a uniform magnetic field of $0.4\text{ T}$. The magnetic field is uniformly reduced to $0.1\text{ T}$ in a time interval of $0.2\text{ s}$. If the electrical resistance of the coil is $5\ \Omega$, calculate: (i) Initial and final magnetic flux linked with the coil. (ii) Magnitude of the induced electromotive force (emf) generated in the coil. (iii) Induced current flowing through the coil. (iv) Total charge flowing through the coil during this time interval. (b) State two methods by which the magnetic flux linked with a loop can be changed.

Part (a): Numerical Solution

Given Data:

• Side of the square coil, $a = 10\text{ cm} = 0.1\text{ m}$
• Area of cross-section, $A = a^2 = (0.1)^2 = 0.01\text{ m}^2 = 10^{-2}\text{ m}^2$
• Number of turns, $N = 500$
• Initial magnetic field, $B_1 = 0.4\text{ T}$
• Final magnetic field, $B_2 = 0.1\text{ T}$
• Time interval, $\Delta t = 0.2\text{ s}$
• Resistance of the coil, $R = 5\ \Omega$
• Angle between field and normal to coil area, $\theta = 0^\circ$ (since coil plane is perpendicular to field)

Step 1: Initial and Final Magnetic Flux ($\Phi_1$ and $\Phi_2$)\nFormula for total magnetic flux linkage:

$$ \Phi = N B A \cos\theta $$

• Initial Flux ($\Phi_1$): $$ \Phi_1 = N B_1 A \cos(0^\circ) = 500 \times 0.4 \times 0.01 \times 1 = 2.0\text{ Wb} $$

• Final Flux ($\Phi_2$): $$ \Phi_2 = N B_2 A \cos(0^\circ) = 500 \times 0.1 \times 0.01 \times 1 = 0.5\text{ Wb} $$

Step 2: Induced Electromotive Force ($e$)\nAccording to Faraday's law of induction:

$$ e = -\frac{\Delta \Phi}{\Delta t} = -\frac{\Phi_2 - \Phi_1}{\Delta t} $$ $$ e = -\frac{0.5 - 2.0}{0.2} = -\frac{-1.5}{0.2} = 7.5\text{ V} $$ $$ \text{Magnitude of induced emf, } |e| = 7.5\text{ V} $$

Step 3: Induced Current ($I$)\nUsing Ohm's law:

$$ I = \frac{|e|}{R} = \frac{7.5}{5} = 1.5\text{ A} $$

Step 4: Total Charge Flowing ($q$)

$$ q = I \times \Delta t = 1.5 \times 0.2 = 0.3\text{ C} $$ (Alternatively, $q = \frac{|\Delta \Phi|}{R} = \frac{1.5}{5} = 0.3\text{ C}$)

Part (b): Methods to Change Magnetic Flux

1. By changing the magnetic field strength ($B$): Increasing or decreasing the strength of the magnetic field passing through the loop.
2. By changing the orientation of the coil ($\theta$): Rotating the coil inside a constant magnetic field so that the angle between field lines and surface vector changes.