(D) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = (B_0 e^{-t})(\pi r^2)$.According to Faraday's law,the induced electromot

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(D) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = (B_0 e^{-t})(\pi r^2)$.According to Faraday's law,the induced electromotive force $(EMF)$ $e$ is given by $e = -\frac{d\phi}{dt}$.$e = -\frac{d}{dt}(B_0 \pi r^2 e^{-t}) = B_0 \pi r^2 e^{-t}$.The electrical power $P$ developed in the resistance $R$ is $P = \frac{e^2}{R}$.$P = \frac{(B_0 \pi r^2 e^{-t})^2}{R} = \frac{B_0^2 \pi^2 r^4 e^{-2t}}{R}$.Right after closing the switch,the time $t = 0$.Substituting $t = 0$ into the expression for power,we get $P = \frac{B_0^2 \pi^2 r^4 e^0}{R} = \frac{B_0^2 \pi^2 r^4}{R}$.

Class 12 Physics Electromagnetic Induction Mix Examples-Electromagnetic Induction

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Shown in the figure is a circular loop of radius $r$ and resistance $R$. $A$ variable magnetic field of induction $B = B_0 e^{-t}$ is established inside the coil. If the key $K$ is closed,the electrical power developed right after closing the switch is equal to

A
$\frac{B_0^2 \pi r^2}{R}$
B
$\frac{B_0 10 r^3}{R}$
C
$\frac{B_0^2 \pi^2 r^4 R}{5}$
D
$\frac{B_0^2 \pi^2 r^4}{R}$

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Mix Examples-Electromagnetic Induction

In the circuit shown below,all the inductors (assumed ideal) and resistors are identical. The current through the resistance on the right is $I$ after the key $K$ has been switched $ON$ for a long time. The currents through the three resistors (in order,from left to right) immediately after the key is switched $OFF$ are

DifficultKVPY 2015

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The figure shown has two coils of wires placed in close proximity. The current in primary coil $P$ is made to vary with time as shown in the graph. Which of the following graphs best represents the variation of the emf induced in the secondary coil $S$?

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$A$ part of a complete circuit is shown in the figure. At some instant,the value of current $I$ is $1\, A$ and it is decreasing at a rate of $10^{2}\, A s^{-1}$. The value of the potential difference $V_{P} - V_{Q}$ (in volts) at that instant is:

MediumJEE MAIN 2020

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$A$ square coil $ABCD$ is placed in the $x-y$ plane with its centre at the origin. $A$ long straight wire,passing through the origin,carries a current in the negative $z$-direction. The current in this wire increases with time. The induced current in the coil is:

Medium

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Assertion: $A$ current continues to flow in a superconducting coil even after the switch is off.
Reason: Superconducting coils show the Meissner effect.

EasyAIIMS 2007

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Shown in the figure is a circular loop of radius $r$ and resistance $R$. $A$ variable magnetic field of induction $B = B_0 e^{-t}$ is established inside the coil. If the key $K$ is closed,the electrical power developed right after closing the switch is equal to

Similar Questions

The figure shown has two coils of wires placed in close proximity. The current in primary coil $P$ is made to vary with time as shown in the graph. Which of the following graphs best represents the variation of the emf induced in the secondary coil $S$?

$A$ part of a complete circuit is shown in the figure. At some instant,the value of current $I$ is $1\, A$ and it is decreasing at a rate of $10^{2}\, A s^{-1}$. The value of the potential difference $V_{P} - V_{Q}$ (in volts) at that instant is:

$A$ square coil $ABCD$ is placed in the $x-y$ plane with its centre at the origin. $A$ long straight wire,passing through the origin,carries a current in the negative $z$-direction. The current in this wire increases with time. The induced current in the coil is:

Assertion: $A$ current continues to flow in a superconducting coil even after the switch is off.Reason: Superconducting coils show the Meissner effect.

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Assertion: $A$ current continues to flow in a superconducting coil even after the switch is off.
Reason: Superconducting coils show the Meissner effect.