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 :
Solution
Flux $\phi$ = $\overrightarrow B .\overrightarrow A$ = BAcos$\omega$t
<br><br>Induced emf = e = $- {{d\phi } \over {dt}}$ = -BA$\omega$(-)sin$\omega$t
<br><br>= BA$\omega$sin$\omega$t
<br><br>e will be maximum at $\omega$t = ${\pi \over 2}$, ${{3\pi } \over 2}$
<br><br>$\Rightarrow$ ${{2\pi } \over T}t$ = ${\pi \over 2}$, ${{3\pi } \over 2}$
<br><br>$\Rightarrow$ t = ${T \over 4}$ or ${3T \over 4}$ i.e. 2.5 s or 7.5 s.
<br><br>For induced emf to be minimum i.e zero.
<br><br>${{2\pi t} \over T}$ = n$\pi$
<br><br>$\Rightarrow$ t = $n{\pi \over 2}$
<br><br>$\Rightarrow$ Induced emf is zero at t = 5 s, 10 s
About this question
Subject: Physics · Chapter: Electromagnetic Induction · Topic: Faraday's Laws
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