Question

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?

• A. \(Ba^{2}\omega\); \(\theta =90^{\circ }\)
• B. \(Ba^{2}\omega ^{2}\); \(\theta =0^{\circ }\)
• C. \(\frac{1}{2}Ba^{2}\omega\); \(\theta =270^{\circ }\)
• D. \(Ba\omega\); \(\theta =180^{\circ }\)

Hint:

Write the flux as \(Ba^2\cos\theta\) and differentiate; the emf is largest when the flux is changing fastest.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The flux through the loop is \(\phi=Ba^{2}\cos\theta\) with \(\theta=\omega t\), the angle between \(\vec{B}\) and the normal.

Step 2: Induced emf
\(e=-\dfrac{d\phi}{dt}=Ba^{2}\omega\sin\omega t\).

Step 3: Maximum value
The emf is largest when \(\sin\theta=1\), that is, first at \(\theta=90^{\circ}\). There the loop plane is parallel to the field and the flux is zero but changing fastest.
So \(e_{max}=Ba^{2}\omega\).

Final Answer:
The maximum emf is \(Ba^{2}\omega\), first at \(\theta=90^{\circ}\). \[ \boxed{\text{(A) }Ba^{2}\omega ;\ \theta =90^{\circ }} \]