Appeared in: 2025 (3 sets)
(a) (i) Define self-inductance of a coil. Derive the expression for the energy required to build up a current $I$ in a coil of self-inductance $L$.
(ii) The currents passing through two inductors of self-inductances $10\text{ mH}$ and $20\text{ mH}$ increase with time at the same rate.
Draw graphs showing the variation of :
(I) the magnitude of emf induced with the rate of change of current in each inductor.
(II) the energy stored in each inductor with the current flowing through it.
OR
(b) (i) Define the term mutual inductance. Deduce the expression for the mutual inductance of two long coaxial solenoids of the same length having different radii and different number of turns.
(ii) The current through an inductor is uniformly increased from zero to $2\text{ A}$ in $40\text{ s}$. An emf of $5\text{ mV}$ is induced during this period. Find the flux linked with the inductor at $t = 10\text{ s}$.
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Solution (a)
(i) Definition and Derivation: [2 Marks]
- Definition: Self-inductance of a coil is defined as the magnetic flux linked with the coil when a unit current flows through it. Alternatively, it is the ratio of the induced emf to the rate of change of current in the coil. ($L = \frac{\phi}{I}$ or $L = \frac{|\varepsilon|}{dI/dt}$). [0.5 Mark]
- Derivation: When current $I$ flows through a coil, the induced emf is $\varepsilon = -L \frac{dI}{dt}$. The work done ($dW$) by an external agent to overcome this induced emf in time $dt$ is: $$dW = |\varepsilon| \cdot I \cdot dt = \left( L \frac{dI}{dt} \right) I \cdot dt = L \cdot I \cdot dI$$ Integrating from $0$ to $I$: $$W = \int_{0}^{I} L \cdot I \cdot dI = L \left[ \frac{I^2}{2} \right]_{0}^{I}$$ $$U = \frac{1}{2} LI^2$$ [1.5 Marks]
(ii) Graphical Representation: [3 Marks]
- (I) Magnitude of emf ($\varepsilon$) vs Rate of change of current ($dI/dt$): Since $\varepsilon = L(dI/dt)$, the graph is a straight line passing through the origin. The slope of the line represents the self-inductance ($L$). The line for $20\text{ mH}$ will have a steeper slope than the $10\text{ mH}$ line. [1.5 Marks]
- (II) Energy stored ($U$) vs Current ($I$): Since $U = \frac{1}{2}LI^2$, the graph is a parabola opening upwards. The curve for $20\text{ mH}$ will rise more sharply than the $10\text{ mH}$ curve. [1.5 Marks]
💡 EL10 Concept Insight: Think of self-inductance like "electrical inertia"—just as a heavy object resists being pushed into motion, an inductor resists any change in the current flowing through it, storing energy in its magnetic field like a compressed spring.
Solution (b)
(i) Mutual Inductance Derivation: [3 Marks]
- Definition: Mutual inductance of two coils is the magnetic flux linked with one coil per unit current flowing through the other coil. [0.5 Mark]
- Derivation: Consider two coaxial solenoids of length $l$, radii $r_1$ and $r_2$ ($r_2 > r_1$), and turns $N_1$ and $N_2$. Magnetic field due to solenoid 2 (carrying current $I_2$): $B_2 = \mu_0 n_2 I_2 = \mu_0 \frac{N_2}{l} I_2$. Flux linked with solenoid 1: $\phi_1 = N_1 B_2 A_1 = N_1 (\mu_0 \frac{N_2}{l} I_2) (\pi r_1^2)$. Since $\phi_1 = M I_2$, we get: $$M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l}$$ [2.5 Marks]
(ii) Numerical Calculation: [2 Marks]