JEE MAIN - Physics (2021 - 27th July Evening Shift - No. 21)
In the given figure the magnetic flux through the loop increases according to the relation $$\phi$$B(t) = 10t2 + 20t, where $$\phi$$B is in milliwebers and t is in seconds.
The magnitude of current through R = 2$$\Omega$$ resistor at t = 5 s is ___________ mA.
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The magnitude of current through R = 2$$\Omega$$ resistor at t = 5 s is ___________ mA.
Answer
60
Explanation
$$\left| \in \right| = {{d\phi } \over {dt}}$$ = 20t + 20 mV
$$\left| i \right| = {{\left| \in \right|} \over R}$$ = 10t + 10 mA
at t = 5
$$\left| i \right|$$ = 60 mA
$$\left| i \right| = {{\left| \in \right|} \over R}$$ = 10t + 10 mA
at t = 5
$$\left| i \right|$$ = 60 mA
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