CBSE Class 12 Physics sample paper 2025-26: question 28

Hints

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Step-by-step solution

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1. Solenoid: $n=\frac{2000}{0.5}=4000$ turns/m; $B={\mu }_{0}nI=4\pi \times 10^{-7}\times 4000\times 5=8\pi \times 10^{-3}$ T

   1 mark

2. When the coil rotates with angular speed $\omega$ (about a diameter ⟂ the solenoid axis), flux $\varphi =N_{c}BA\cos \omega t$, so $\epsilon =-\frac{d\varphi }{dt}=N_{c}BA\omega \sin \omega t$ and ${\epsilon }_{max}=N_{c}BA\omega$.

   1 mark

3. ${\epsilon }_{max}=50\times 8\pi \times 10^{-3}\times \frac{5}{\pi }\times 10^{-4}\times \omega =2\times 10^{-4}\omega$ V. Since $\omega$ is not given, ${\epsilon }_{max}=2\times 10^{-4}\omega$ volt (e.g. $2\times 10^{-4}$ V for $\omega =1$ rad/s).

   1 mark

Answer: ${\epsilon }_{max}=N_{c}BA\omega =2\times 10^{-4}\omega$ V (B = 8π×10⁻³ T)

Revise the chapter

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