CBSE · Class 12 · Physics · Electromagnetic Induction(a) What is meant by self-induction? Derive an expression for the self-inductance of a long air-cored solenoid. State the factors on which the self-inductance of a solenoid depends. (b) A magnetic flux linked with a coil of $200$ turns changes from $8 \times 10^{-4} \text{ Wb}$ to $2 \times 10^{-4} \text{ Wb}$ in $0.02 \text{ seconds}$. Calculate the magnitude of induced electromotive force (emf) across the coil.

Part (a): Self-Induction and Derivation for a Long Solenoid

Definition of Self-Induction:\nSelf-induction is the phenomenon by which an electromotive force (emf) is induced in a circuit or coil due to a change in the electric current passing through the same coil. According to Faraday's law of electromagnetic induction, whenever the electric current flowing through a coil changes, the magnetic flux linked with the coil also changes. This changing magnetic flux induces an opposing emf in the coil itself, which opposes the growth or decay of current.

Derivation of Self-Inductance of a Long Solenoid:\nConsider a long air-cored solenoid of length $L$, uniform cross-sectional area $A$, and total number of turns $N$.

\nThe number of turns per unit length is given by: $$n = \frac{N}{L}$$ \nWhen a current $I$ flows through the solenoid, a uniform magnetic field $B$ is produced inside it along its axis: $$B = \mu_0 n I = \frac{\mu_0 N I}{L}$$ \nThe magnetic flux $\phi_B$ linked with each individual turn of the solenoid is: $$\phi_B = B \cdot A = \left(\frac{\mu_0 N I}{L}\right) A$$ \nTherefore, the total magnetic flux linkage ($\Phi_{\text{total}}$) for all $N$ turns of the solenoid is: $$\Phi_{\text{total}} = N \cdot \phi_B = N \cdot \left(\frac{\mu_0 N I A}{L}\right) = \frac{\mu_0 N^2 A I}{L}$$ \nBy the definition of self-inductance ($L_s$): $$\Phi_{\text{total}} = L_s \cdot I$$ \nEquating both expressions for total magnetic flux linkage: $$L_s \cdot I = \frac{\mu_0 N^2 A I}{L}$$ $$L_s = \frac{\mu_0 N^2 A}{L}$$ \nIf the solenoid is wound over a magnetic core of relative permeability $\mu_r$, the self-inductance is: $$L_s = \frac{\mu_0 \mu_r N^2 A}{L}$$

Factors Affecting Self-Inductance:

1. Number of Turns ($N$): Self-inductance is directly proportional to the square of the number of turns ($L_s \propto N^2$).
2. Area of Cross-Section ($A$): Self-inductance is directly proportional to the cross-sectional area of the solenoid ($L_s \propto A$).
3. Length of Solenoid ($L$): Self-inductance is inversely proportional to the length of the solenoid ($L_s \propto \frac{1}{L}$).
4. Permeability of Core Material ($\mu_r$): Inserting a material with high relative magnetic permeability (such as soft iron) increases the self-inductance significantly.

Part (b): Numerical Solution

Given Data:

• Number of turns, $N = 200$
• Initial magnetic flux, $\phi_1 = 8 \times 10^{-4} \text{ Wb}$
• Final magnetic flux, $\phi_2 = 2 \times 10^{-4} \text{ Wb}$
• Time interval, $\Delta t = 0.02 \text{ s}$

Formula:\nAccording to Faraday's law of electromagnetic induction, the magnitude of induced emf $e$ is: $$e = N \left| \frac{\Delta \phi}{\Delta t} \right| = N \left| \frac{\phi_2 - \phi_1}{\Delta t} \right|$$

Calculation: $$\Delta \phi = \phi_2 - \phi_1 = (2 \times 10^{-4}) - (8 \times 10^{-4}) = -6 \times 10^{-4} \text{ Wb}$$ $$\left| \Delta \phi \right| = 6 \times 10^{-4} \text{ Wb}$$ \nSubstituting values into the formula: $$e = 200 \times \frac{6 \times 10^{-4}}{0.02}$$ $$e = 200 \times \frac{6 \times 10^{-4}}{2 \times 10^{-2}}$$ $$e = 200 \times (3 \times 10^{-2})$$ $$e = 6 \text{ V}$$

Answer:\nThe magnitude of the induced electromotive force across the coil is $6 \text{ V}$.