CBSE 2025, Set 1

Question

(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 A in 40 s. An emf of 5 mV is induced during this period. Find the flux linked with the inductor at $t = 10$ s.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Mutual inductance — definition and derivation:** **Definition:** Mutual inductance of two coils is defined as the emf induced in one coil per unit rate of change of current in the other coil. Alternatively, it is the magnetic flux linked with one coil when unit current flows in the other. $$M = \frac{N_1\phi_1}{I_2} = \frac{|e_1|}{dI_2/dt}$$ **Derivation for two coaxial solenoids:** ![Two coaxial solenoids](https://www.edvaya.com/images/cbse/physics/2025/set1/q32b-coaxial-solenoids-mutual-inductance.webp) Consider two long coaxial solenoids $S_1$ (inner, radius $r_1$, $n_1$ turns/unit length) and $S_2$ (outer, radius $r_2$, $n_2$ turns/unit length), both of length $l$. When current $I_2$ is set up in outer solenoid $S_2$, the magnetic field inside $S_2$: $$B_2 = \mu_0 n_2 I_2$$ The flux through the inner solenoid $S_1$ due to $B_2$: $$\phi_1 = B_2 \cdot \pi r_1^2 = \mu_0 n_2 I_2 \pi r_1^2$$ Total flux linkage through $S_1$ ($N_1 = n_1 l$ total turns): $$N_1\phi_1 = (n_1 l)(\mu_0 n_2 I_2 \pi r_1^2) = \mu_0 n_1 n_2 \pi r_1^2 l \cdot I_2$$ By definition: $M_{12} = \dfrac{N_1\phi_1}{I_2}$: $$\boxed{M = \mu_0 n_1 n_2 \pi r_1^2 l}$$ This result equals $M_{21}$ (the flux in $S_2$ due to unit current in $S_1$), confirming the reciprocity theorem. --- **(ii) Flux linked at t = 10 s:** **Step 1:** Find self-inductance L from given data: $$|e| = L\frac{dI}{dt} \implies L = \frac{|e|}{dI/dt} = \frac{5 	imes 10^{-3}}{2/40} = \frac{5 	imes 10^{-3}}{0\cdot05} = 0\cdot1 	ext{ H}$$ **Step 2:** Find current at t = 10 s: Since current increases uniformly from 0 to 2 A in 40 s: $$I_{10} = \frac{2}{40} 	imes 10 = 0\cdot5 	ext{ A}$$ **Step 3:** Flux linked: $$\phi = LI = 0\cdot1 	imes 0\cdot5 = \boxed{0\cdot05 	ext{ Wb}}$$

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Physics > Magnetism Induction And Alternating Current > Electromagnetic Induction

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