(C) The magnetic flux through the coil is given by $\phi = 2t^2 + 3t - 2$. According to Faraday's law of electromagnetic induction,the induced electro

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(C) The magnetic flux through the coil is given by $\phi = 2t^2 + 3t - 2$. According to Faraday's law of electromagnetic induction,the induced electromotive force $(e)$ is given by $e = -\frac{d\phi}{dt}$. $e = -\frac{d}{dt}(2t^2 + 3t - 2) = -(4t + 3)$. The magnitude of the induced $EMF$ is $|e| = 4t + 3$. At time $t = 3 \ s$,the induced $EMF$ is $|e| = 4(3) + 3 = 12 + 3 = 15 \ V$. The induced current in the coil is $I = \frac{|e|}{R} = \frac{15 \ V}{7.5 \ \Omega} = 2 \ A$. The induced power $(P)$ in the coil is given by $P = I^2 R = (2)^2 \times 7.5 = 4 \times 7.5 = 30 \ W$.

Class 12 Physics Electromagnetic Induction Mix Examples-Electromagnetic Induction

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The plane of a circular coil of resistance $7.5 \ \Omega$ is placed perpendicular to a uniform magnetic field. The flux $\phi$ (in weber) through the coil varies with time $t$ (in second) as $\phi = 2t^2 + 3t - 2$. The induced power in the coil at time $t = 3 \ s$ is (in $W$)

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$7.5$
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$15$
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$30$
D
$20$

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$A$ coil is placed in a time-varying magnetic field. If the number of turns in the coil were to be halved and the radius of the wire doubled,the electrical power dissipated due to the current induced in the coil would be: (Assume the coil to be short-circuited.)

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The switch $S$ of the circuit shown in the figure is closed at $t = 0$. If $e$ denotes the induced emf in the inductor $L$ and $i$ denotes the current flowing through the circuit at time $t$,which of the following graphs correctly represents the variation of $e$ with time $t$?

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$A$ conducting wire frame is placed in a magnetic field which is directed into the paper. The magnetic field is increasing at a constant rate. The directions of induced currents in wires $AB$ and $CD$ are

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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

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$A$ current $I = 10 \ A$ is passed through the part of a circuit shown in the figure. What will be the potential difference between $A$ and $B$ when $I$ is decreased at a constant rate of $10^2 \ A \ s^{-1}$ (in $V$)?

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The plane of a circular coil of resistance $7.5 \ \Omega$ is placed perpendicular to a uniform magnetic field. The flux $\phi$ (in weber) through the coil varies with time $t$ (in second) as $\phi = 2t^2 + 3t - 2$. The induced power in the coil at time $t = 3 \ s$ is (in $W$)

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$A$ coil is placed in a time-varying magnetic field. If the number of turns in the coil were to be halved and the radius of the wire doubled,the electrical power dissipated due to the current induced in the coil would be: (Assume the coil to be short-circuited.)

The switch $S$ of the circuit shown in the figure is closed at $t = 0$. If $e$ denotes the induced emf in the inductor $L$ and $i$ denotes the current flowing through the circuit at time $t$,which of the following graphs correctly represents the variation of $e$ with time $t$?

$A$ conducting wire frame is placed in a magnetic field which is directed into the paper. The magnetic field is increasing at a constant rate. The directions of induced currents in wires $AB$ and $CD$ are

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

$A$ current $I = 10 \ A$ is passed through the part of a circuit shown in the figure. What will be the potential difference between $A$ and $B$ when $I$ is decreased at a constant rate of $10^2 \ A \ s^{-1}$ (in $V$)?

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