CBSE Class 12 Physics 2026: question 18

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Step-by-step solution

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1. At $t=0$ the plane of the coil is parallel to $\vec{B}$, so its normal is perpendicular to $\vec{B}$ and the flux is zero. After time t the normal has turned through $\omega t$ from this position, so the angle between normal and $\vec{B}$ is $(90^{\circ }-\omega t)$:
   $\varphi =NBA\cos (90^{\circ }-\omega t)=NBA\sin \omega t$

2. (a) Plot of $\varphi$ vs $\omega t$: a sine curve starting at zero, rising to maximum $+NBA$ at $\omega t=\pi /2$, zero at $\pi$, minimum $-NBA$ at $3\pi /2$, zero at $2\pi$.

   1 mark

3. (b) $\epsilon =-\frac{d\varphi }{dt}=-NBA\omega \cos \omega t$.
   Plot of $\epsilon$ vs $\omega t$: a (negative) cosine curve: $-NBA\omega$ at $\omega t=0$, zero at $\pi /2$, $+NBA\omega$ at $\pi$, zero at $3\pi /2$, $-NBA\omega$ at $2\pi$. The emf is maximum in magnitude when the flux is zero, and zero when the flux is maximum (90° phase difference).

   1 mark

Answer: φ = NBA sin ωt (sine curve from zero); ε = −NBAω cos ωt (cosine curve, maximum magnitude at ωt = 0)

Revise the chapter

• Chapter: Electromagnetic Induction (notes, formulas, where marks are lost and every past question)