(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)

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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).

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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Motional EMI (Induced Parameter)

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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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$A$ conducting rod $AB$ of length $l = 1\,m$ is moving with a velocity $v = 4\,m/s$. The velocity vector makes an angle of $30^o$ with the length of the rod. $A$ uniform magnetic field $B = 2\,T$ exists in a direction perpendicular to the plane of motion. Then:

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$A$ horizontal straight wire $20 \; m$ long extending from east to west is falling with a speed of $5.0 \; m/s$ at right angles to the horizontal component of the earth's magnetic field $0.30 \times 10^{-4} \; Wb/m^2$. The instantaneous value of the emf induced in the wire will be ......... $mV$.

MediumAIEEE 2011

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$A$ conducting ring of radius $a$ is rotated about a point $O$ on its periphery as shown in the figure in a plane perpendicular to a uniform magnetic field $B$ which exists everywhere. The rotational velocity is $\omega$. Choose the correct statement$(s)$ related to the magnitude of potential differences.

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$A$ boy is playing with the empty rim of a cycle wheel of radius $40 \, cm$ by rolling it along a horizontal road towards north with an angular speed of $20 \, rad \, s^{-1}$. Considering the effect of the magnetic field of the Earth, the e.m.f. induced in the rim is (Horizontal component of Earth's magnetic field $= 0.26 \, G$)

MediumTS EAMCET 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).

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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.

$A$ conducting rod $AB$ of length $l = 1\,m$ is moving with a velocity $v = 4\,m/s$. The velocity vector makes an angle of $30^o$ with the length of the rod. $A$ uniform magnetic field $B = 2\,T$ exists in a direction perpendicular to the plane of motion. Then:

$A$ horizontal straight wire $20 \; m$ long extending from east to west is falling with a speed of $5.0 \; m/s$ at right angles to the horizontal component of the earth's magnetic field $0.30 \times 10^{-4} \; Wb/m^2$. The instantaneous value of the emf induced in the wire will be ......... $mV$.

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