Question

A planar loops 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 :

Answer 1

Correct option is C. $$2.5 s$$ and $$5.0 s$$
Time period, $$T = 10 s$$

Ref. image I

Axis of rotation $$\rightarrow $$ y-axis.

$$\vec{B} = B(-\hat{k})$$

ref. image II

at some time t, area vector makes angle '$$\theta$$' with $$\vec{B}$$.

flux, $$\phi = \vec{B}. \vec{A}$$

$$= BA \cos \theta$$

emf, $$\varepsilon= \left|\dfrac{-d \phi}{dt} \right| = \left(BA \sin \theta \dfrac{d \theta}{dt}\right)$$

$$= BA \omega \sin \theta$$

When $$\theta = \dfrac{\pi}{2} , \varepsilon = \varepsilon_{max}$$

So, $$\theta = n \pi + \dfrac{\pi}{2} $$ n = integes

When $$\theta = 0, \varepsilon = \varepsilon_{min} = 0$$

So, $$\theta = n \pi$$, n = integes

for $$\varepsilon = \varepsilon_{max}, \theta = \omega t = n \pi + \dfrac{\pi}{2}$$

$$\Rightarrow t = \dfrac{n \pi}{\omega} + \dfrac{\pi}{2 \omega} = \dfrac{n \pi}{\left(\dfrac{2 \pi}{T} \right)} + \dfrac{\pi}{2 \times \left(\dfrac{2 \pi}{T} \right)}$$

$$t = \left(\dfrac{n \pi}{\dfrac{2 \pi}{T}} \right) + \dfrac{\pi}{2 \left(\dfrac{2 \pi}{T} \right)}$$

$$= \dfrac{n T}{2} + \dfrac{T}{4} = \dfrac{n \times 10}{2} + \dfrac{10}{4}$$

$$= 5n + 2.5$$

$$= 2.5 s, 5 + 2.5 , 5 \times 2 + 2.5 ...$$

$$= 2.5 s, 7.5 s, 12.5 s....$$

for $$\varepsilon = \varepsilon_{min}, \theta = \omega t = n \pi$$

$$\Rightarrow t = \dfrac{n \pi}{\omega} = \dfrac{n \pi}{\left(\dfrac{2 \pi}{T} \right)} = \dfrac{n T}{2}$$

$$\Rightarrow t = \dfrac{n \times 10}{2} = 5n$$

$$= 0 s, 5s, 10s, ...$$

So, possible option is option (C).