(C) The induced $emf$ $(E)$ in a rotating disc is given by the formula: $E = \frac{1}{2} B \omega R^2$.Given:Radius $(R)$ = $0.1 \ m$Magnetic field $(

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(C) The induced $emf$ $(E)$ in a rotating disc is given by the formula: $E = \frac{1}{2} B \omega R^2$.Given:Radius $(R)$ = $0.1 \ m$Magnetic field $(B)$ = $0.1 \ T$Frequency $(f)$ = $10 \ rev/s$Angular velocity $(\omega)$ = $2\pi f = 2 \times \pi \times 10 = 20\pi \ rad/s$.Substituting the values:$E = \frac{1}{2} \times 0.1 \times (20\pi) \times (0.1)^2$$E = 0.1 \times 10\pi \times 0.01$$E = 1 \times \pi \times 0.01 = 0.01\pi \ V$Converting to millivolts $(mV)$:$E = 0.01\pi \times 1000 \ mV = 10\pi \ mV$.Therefore,the correct option is $C$.

Class 12 Physics Electromagnetic Induction Motional EMI (Induced Parameter)

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$A$ circular disc of radius $0.1 \ m$ is rotating about its axis at a frequency of $10 \ rev/s$ in a uniform magnetic field of $0.1 \ T$ directed perpendicular to the disc. Calculate the induced $emf$ between the center and the rim of the disc.

A
$\frac{\pi}{10} \ V$
B
$\frac{2\pi}{10} \ V$
C
$10\pi \ mV$
D
$20\pi \ mV$

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Electromagnetic Induction

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Motional EMI (Induced Parameter)

$A$ wire frame $PQRSTU$ is moving horizontally with velocity $v$ in a uniform magnetic field $B$ acting perpendicular to its plane as shown in the figure. Choose the $INCORRECT$ statement.

Medium

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$A$ metal conductor of length $1\;m$ rotates vertically about one of its ends at an angular velocity of $5\;rad/s$. If the horizontal component of the Earth's magnetic field is $0.2 \times 10^{-4}\;T$,then the $e.m.f.$ developed between the two ends of the conductor is:

MediumAIEEE 2004

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The magnetic field in a region is given by $\vec B = B_0(1 + \frac{x}{a})\hat k$. $A$ square loop of edge length $d$ is placed with its edges along the $x$ and $y$ axes. The loop is moved with a constant velocity $\vec V = V_0\hat i$. The $emf$ induced in the loop is:

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$A$ rod of $10 \ cm$ length is moving perpendicular to a uniform magnetic field of intensity $5 \times 10^{-4} \ Wb/m^2$. If the acceleration of the rod is $5 \ m/s^2$, then the rate of increase of induced $emf$ is . . . . . . .

EasyGUJCET 2015

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$A$ wheel of $20$ metallic spokes, each $40 \text{ cm}$ long, is rotated with a speed of $180 \text{ rev/min}$ in a plane normal to the horizontal component of Earth's magnetic field $H_{e}$ at a place. If $H_{e} = 0.4 \text{ G}$ at that place, the induced emf between the axle and the rim of the wheel is:

DifficultTS EAMCET 2022

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$A$ circular disc of radius $0.1 \ m$ is rotating about its axis at a frequency of $10 \ rev/s$ in a uniform magnetic field of $0.1 \ T$ directed perpendicular to the disc. Calculate the induced $emf$ between the center and the rim of the disc.

Similar Questions

$A$ wire frame $PQRSTU$ is moving horizontally with velocity $v$ in a uniform magnetic field $B$ acting perpendicular to its plane as shown in the figure. Choose the $INCORRECT$ statement.

$A$ metal conductor of length $1\;m$ rotates vertically about one of its ends at an angular velocity of $5\;rad/s$. If the horizontal component of the Earth's magnetic field is $0.2 \times 10^{-4}\;T$,then the $e.m.f.$ developed between the two ends of the conductor is:

The magnetic field in a region is given by $\vec B = B_0(1 + \frac{x}{a})\hat k$. $A$ square loop of edge length $d$ is placed with its edges along the $x$ and $y$ axes. The loop is moved with a constant velocity $\vec V = V_0\hat i$. The $emf$ induced in the loop is:

$A$ rod of $10 \ cm$ length is moving perpendicular to a uniform magnetic field of intensity $5 \times 10^{-4} \ Wb/m^2$. If the acceleration of the rod is $5 \ m/s^2$, then the rate of increase of induced $emf$ is . . . . . . .

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