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 :

A
$$5.0 s$$ and $$7.5 s$$
B
$$2.5 s$$ and $$7.5 s$$
C
$$2.5 s$$ and $$5.0 s$$
D
$$5.0 s$$ and $$10.0 s$$
Solution
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#### Correct option is C. $$2.5 s$$ and $$5.0 s$$Time period, $$T = 10 s$$Ref. image IAxis of rotation $$\rightarrow$$ y-axis.$$\vec{B} = B(-\hat{k})$$ref. image IIat 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 = integesfor $$\varepsilon = \varepsilon_{max}, \theta = \omega t = n \pi + \dfrac{\pi}{2}$$$$\Rightarrow t = \dfrac{n \pi}{\omega} + \dfrac{\pi}{2 \omega} = \dfrac{n \pi}{\left(\dfrac{2 \pi}{T} \right)} + \dfrac{\pi}{2 \times \left(\dfrac{2 \pi}{T} \right)}$$$$t = \left(\dfrac{n \pi}{\dfrac{2 \pi}{T}} \right) + \dfrac{\pi}{2 \left(\dfrac{2 \pi}{T} \right)}$$$$= \dfrac{n T}{2} + \dfrac{T}{4} = \dfrac{n \times 10}{2} + \dfrac{10}{4}$$$$= 5n + 2.5$$$$= 2.5 s, 5 + 2.5 , 5 \times 2 + 2.5 ...$$$$= 2.5 s, 7.5 s, 12.5 s....$$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).

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