Question

A planar loop of wire rotates in a uniform magnetic field. Initially at t = 0, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of 10 s about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at :

• A. 2.5 s and 7.5 s
• B. 5.0 s and 10.0 s
• C. 5.0 s and 7.5 s
• D. 2.5 s and 5.0 s

Hint:

Flux \(\phi=BA\cos\omega t\) (loop starts perpendicular to B), so emf \(=BA\omega\sin\omega t\). Find when \(\sin\omega t\) is 1 and 0, with \(\omega=2\pi/T\).

Answer 1

The Correct Option is D

Step 1: Loop plane is perpendicular to B at t = 0, so \(\phi=BA\cos\omega t\) with \(\omega=\dfrac{2\pi}{10}\) rad/s.
Step 2: \(\varepsilon=-\dfrac{d\phi}{dt}=BA\omega\sin\omega t\).
Step 3: Maximum emf when \(\sin\omega t=1\): \(\omega t=\dfrac\pi2\Rightarrow t=\dfrac T4=2.5\) s.
Step 4: Minimum (zero) emf when \(\sin\omega t=0\): \(\omega t=\pi\Rightarrow t=\dfrac T2=5.0\) s.
Answer: (D) 2.5 s and 5.0 s