Q181 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
A square loop (side $a$) rotates in a uniform magnetic field $B$ perpendicular to the rotation axis. What is the maximum induced EMF ($e_{\text{max}}$) and the first angle $\theta$ (between magnetic field $B$ and the loop normal) where it occurs?
- Correct answerA$Ba^2\omega$; $\theta = 90^\circ$
- B$Ba^2\omega^2$; $\theta = 0^\circ$
- C$\dfrac{1}{2}Ba^2\omega$; $\theta = 270^\circ$
- D$Ba\omega$; $\theta = 180^\circ$
Solution
The magnetic flux through the loop at any instant is given by $\Phi = \vec{B} \cdot \vec{A} = B A \cos\theta$, where $\theta$ is the angle between the magnetic field and the normal to the loop area.
Since the loop rotates with an angular velocity $\omega$, we have $\theta = \omega t$.
The area of the square loop is $A = a^2$.
The induced EMF is given by Faraday's law of induction:
$e = -\dfrac{d\Phi}{dt} = -\dfrac{d}{dt}(B a^2 \cos\theta) = B a^2 \omega \sin\theta$
The maximum induced EMF occurs when $\sin\theta = 1$.
The first angle where this occurs is $\theta = 90^\circ$.
The maximum induced EMF is $e_{\text{max}} = B a^2 \omega$.
Answer: $Ba^2\omega$; $\theta = 90^\circ$
Since the loop rotates with an angular velocity $\omega$, we have $\theta = \omega t$.
The area of the square loop is $A = a^2$.
The induced EMF is given by Faraday's law of induction:
$e = -\dfrac{d\Phi}{dt} = -\dfrac{d}{dt}(B a^2 \cos\theta) = B a^2 \omega \sin\theta$
The maximum induced EMF occurs when $\sin\theta = 1$.
The first angle where this occurs is $\theta = 90^\circ$.
The maximum induced EMF is $e_{\text{max}} = B a^2 \omega$.
Answer: $Ba^2\omega$; $\theta = 90^\circ$
Hence, the correct answer is option: 'A'.