A rectangular coil of $N=100$ turns has length $l=0.15\text{m}$ and width $w=0.10\text{m}$ (area $A=0.015{\text{m}}^2$). It rotates with constant angular speed $\omega =200{\text{rads}}^{-1}$ about an axis that makes an angle $\theta =30^{\circ }$ with a uniform magnetic field $B=0.40\text{T}$. (i) Write the expression for the instantaneous induced emf. (ii) Calculate its maximum value.

Step-by-step solution

1. The magnetic flux through the coil is $\Phi =NBA\cos \alpha$, where $\alpha$ is the angle between the coil normal and $B$.
2. Because the rotation axis makes an angle $\theta$ with $B$, the normal makes an angle $\alpha (t)=\theta +\omega t$; using the identity $\cos (\theta +\omega t)=\cos \theta \cos (\omega t)-\sin \theta \sin (\omega t)$, the time‑varying part that produces emf is the $\cos \theta \cos (\omega t)$ term.
3. Hence $\Phi (t)=NBA\cos \theta \cos (\omega t)$.
4. Apply Faraday’s law: $\epsilon =-\frac{d\Phi }{dt}=NBA\cos \theta \omega \sin (\omega t)$.
5. Thus the instantaneous emf is $\epsilon (t)=NBA\omega \cos \theta \sin (\omega t)$.
6. The maximum magnitude occurs when $|\sin (\omega t)|=1$: ${\epsilon }_{\text{max}}=NBA\omega \cos \theta$.
7. Insert the numbers: $N=100$, $B=0.40\text{T}$, $A=0.015{\text{m}}^2$, $\omega =200{\text{rads}}^{-1}$, $\cos 30^{\circ }=\sqrt{3}/2\approx 0.866$.
8. Compute: $NBA=100\times 0.40\times 0.015=0.6$; then $0.6\times 200=120$; finally $120\times 0.866\approx 103.9\text{V}$.
9. Therefore ${\epsilon }_{\text{max}}\approx 1.04\times 10^2\text{V}$.

Concept tested: Rotating coil (alternating emf)

Published 16 September 2026 · How this solution was prepared and checked