(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 of induction,the induc

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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 of induction,the induced electromotive force $(emf)$ is $E = -\frac{d\phi}{dt}$.$E = -\frac{d}{dt} (B_0 \pi r^2 e^{-t}) = -\pi r^2 B_0 \frac{d}{dt}(e^{-t}) = \pi r^2 B_0 e^{-t}$.At the instant the key $(K)$ is closed,$t = 0$.Therefore,the induced $emf$ at $t = 0$ is $E_0 = \pi r^2 B_0 e^0 = \pi r^2 B_0$.The electrical power $P$ developed in the resistor $R$ is given by $P = \frac{E_0^2}{R}$.Substituting the value of $E_0$,we get $P = \frac{(\pi r^2 B_0)^2}{R} = \frac{B_0^2 \pi^2 r^4}{R}$.

Class 12 Physics Electromagnetic Induction Faraday's and Lenz's Law

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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
$\frac{B_0^2 \pi r^3}{R}$
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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Electromagnetic Induction

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Faraday's and Lenz's Law

$A$ circular coil is placed near a current-carrying conductor, both lying on the plane of the paper. The current is flowing through the conductor in such a way that the induced current in the loop is clockwise, as shown in the figure. The current in the wire is,

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$A$ magnetic field of flux density $1.0 \,Wb \,m^{-2}$ acts normal to a $80$ turn coil of $0.01 \,m^2$ area. If this coil is removed from the field in $0.2 \,s$, then the emf induced in it is (in $\,V$)

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Magnetic flux passing through a coil is initially $4 \times 10^{-4} \ Wb$. It reduces to $10 \%$ of its original value in $t$ seconds. If the induced emf is $0.72 \ mV$,then $t$ in seconds is:

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$A$ circular loop made of thin copper wire of mass $m$ is placed in a uniform magnetic field such that the plane of the loop is perpendicular to the magnetic field. If $d$ and $\rho$ are the density and resistivity of copper respectively and the magnetic field varies at a constant rate of $\frac{dB}{dt}$,then the induced current in the loop is . . . . . .

MediumAP EAMCET 2017

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$A$ circular coil of radius $10 \; cm$, $500$ turns, and resistance $2 \; \Omega$ is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through $180^{\circ}$ in $0.25 \; s$. The induced e.m.f. in the coil is (Take $B_H = 3.0 \times 10^{-5} \; T$).

MediumNEET 2017

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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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$A$ circular coil is placed near a current-carrying conductor, both lying on the plane of the paper. The current is flowing through the conductor in such a way that the induced current in the loop is clockwise, as shown in the figure. The current in the wire is,

$A$ magnetic field of flux density $1.0 \,Wb \,m^{-2}$ acts normal to a $80$ turn coil of $0.01 \,m^2$ area. If this coil is removed from the field in $0.2 \,s$, then the emf induced in it is (in $\,V$)

Magnetic flux passing through a coil is initially $4 \times 10^{-4} \ Wb$. It reduces to $10 \%$ of its original value in $t$ seconds. If the induced emf is $0.72 \ mV$,then $t$ in seconds is:

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