(A) The magnetic flux $\phi$ through the coil at time $t$ is given by $\phi = NBA \cos(\omega t)$, where $\omega = 2 \pi v$ is the angular frequency.A

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$NBA(2 \pi v) \sin (2 \pi v t)$

(A) The magnetic flux $\phi$ through the coil at time $t$ is given by $\phi = NBA \cos(\omega t)$, where $\omega = 2 \pi v$ is the angular frequency.According to Faraday's law of induction, the induced $EMF$ $e$ is given by $e = -\frac{d\phi}{dt}$.Substituting the expression for $\phi$: $e = -\frac{d}{dt} [NBA \cos(\omega t)] = -NBA \frac{d}{dt} [\cos(\omega t)]$.Using the chain rule, $e = -NBA [-\omega \sin(\omega t)] = NBA \omega \sin(\omega t)$.Substituting $\omega = 2 \pi v$, we get $e = NBA(2 \pi v) \sin(2 \pi v t)$.

Class 12 Physics Electromagnetic Induction Motional EMI (Induced Parameter)

Easy

In an $AC$ generator, if a coil of $N$ turns and area $A$ is rotated at $v$ revolutions per second in a uniform magnetic field $B$, then the motional $EMF$ produced is equal to (At $t=0$ $s$, the coil is perpendicular to the field).

AP EAMCET 2023 All AP EAMCET

A
$NBA(2 \pi v) \sin (2 \pi v t)$
B
$NBA^2(2 \pi v) \sin (2 \pi v t)$
C
$N^2 B^2 A^2(2 \pi v) \sin (2 \pi v t)$
D
$NBA(4 \pi v) \sin (2 \pi v t)$

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Electromagnetic Induction

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Motional EMI (Induced Parameter)

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$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with uniform speed $v = 20\, m/s$ in a uniform magnetic field $B = 4.0\, T$ directed into the paper. $A$ capacitor of capacity $C = 10\, \mu F$ is connected as shown in the figure. Then:

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$A$ conducting rod of length $\ell$ moves with a velocity $V$ in a uniform magnetic field $B$ perpendicular to the plane of the rails. $A$ capacitor of capacitance $C$ is connected across the rails as shown. Find the charge on the capacitor due to the induced emf in the rod.

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$A$ copper disc of radius $0.1 \ m$ rotates about an axis passing through its centre and perpendicular to its plane with $10 \ \text{revolutions per second}$ in a uniform transverse magnetic field of $0.1 \ T$. The emf induced across the radius of the disc is

DifficultTS EAMCET 2023

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$A$ conducting bar moves on two conducting rails as shown in the figure. $A$ constant magnetic field $B$ exists into the page. The bar starts to move from the vertex at time $t=0$ with a constant velocity $v$. If the induced $\text{EMF}$ is $E \propto t^n$,then the value of $n$ is . . . . . . .

MediumJEE MAIN 2025

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$A$ conducting wire of length $2500 \ m$ is kept in the east-west direction at a height of $10 \ m$ from the ground. If it falls freely to the ground,the induced current in the wire is (Resistance of wire $= 25 \sqrt{2} \ \Omega$,acceleration due to gravity $g = 10 \ m/s^2$,$B_H = 2 \times 10^{-5} \ T$). (in $A$)

DifficultMHT CET 2023

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In an $AC$ generator, if a coil of $N$ turns and area $A$ is rotated at $v$ revolutions per second in a uniform magnetic field $B$, then the motional $EMF$ produced is equal to (At $t=0$ $s$, the coil is perpendicular to the field).

Similar Questions

$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with uniform speed $v = 20\, m/s$ in a uniform magnetic field $B = 4.0\, T$ directed into the paper. $A$ capacitor of capacity $C = 10\, \mu F$ is connected as shown in the figure. Then:

$A$ conducting rod of length $\ell$ moves with a velocity $V$ in a uniform magnetic field $B$ perpendicular to the plane of the rails. $A$ capacitor of capacitance $C$ is connected across the rails as shown. Find the charge on the capacitor due to the induced emf in the rod.

$A$ copper disc of radius $0.1 \ m$ rotates about an axis passing through its centre and perpendicular to its plane with $10 \ \text{revolutions per second}$ in a uniform transverse magnetic field of $0.1 \ T$. The emf induced across the radius of the disc is

$A$ conducting bar moves on two conducting rails as shown in the figure. $A$ constant magnetic field $B$ exists into the page. The bar starts to move from the vertex at time $t=0$ with a constant velocity $v$. If the induced $\text{EMF}$ is $E \propto t^n$,then the value of $n$ is . . . . . . .

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