JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 16)
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 :
2.5 s and 7.5 s
5.0 s and 10.0 s
5.0 s and 7.5 s
2.5 s and 5.0 s
Explanation
Flux $$\phi $$ = $$\overrightarrow B .\overrightarrow A $$ = BAcos$$\omega $$t
Induced emf = e = $$ - {{d\phi } \over {dt}}$$ = -BA$$\omega $$(-)sin$$\omega $$t
= BA$$\omega $$sin$$\omega $$t
e will be maximum at $$\omega $$t = $${\pi \over 2}$$, $${{3\pi } \over 2}$$
$$ \Rightarrow $$ $${{2\pi } \over T}t$$ = $${\pi \over 2}$$, $${{3\pi } \over 2}$$
$$ \Rightarrow $$ t = $${T \over 4}$$ or $${3T \over 4}$$ i.e. 2.5 s or 7.5 s.
For induced emf to be minimum i.e zero.
$${{2\pi t} \over T}$$ = n$$\pi $$
$$ \Rightarrow $$ t = $$n{\pi \over 2}$$
$$ \Rightarrow $$ Induced emf is zero at t = 5 s, 10 s
Induced emf = e = $$ - {{d\phi } \over {dt}}$$ = -BA$$\omega $$(-)sin$$\omega $$t
= BA$$\omega $$sin$$\omega $$t
e will be maximum at $$\omega $$t = $${\pi \over 2}$$, $${{3\pi } \over 2}$$
$$ \Rightarrow $$ $${{2\pi } \over T}t$$ = $${\pi \over 2}$$, $${{3\pi } \over 2}$$
$$ \Rightarrow $$ t = $${T \over 4}$$ or $${3T \over 4}$$ i.e. 2.5 s or 7.5 s.
For induced emf to be minimum i.e zero.
$${{2\pi t} \over T}$$ = n$$\pi $$
$$ \Rightarrow $$ t = $$n{\pi \over 2}$$
$$ \Rightarrow $$ Induced emf is zero at t = 5 s, 10 s
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