(B) Given: Length of the rod $l = 2 \, m$, speed $v = 5 \, ms^{-1}$, magnetic field $B = 0.04 \, T$, and resistance $R = 3 \, \Omega$.The motional ele

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(B) Given: Length of the rod $l = 2 \, m$, speed $v = 5 \, ms^{-1}$, magnetic field $B = 0.04 \, T$, and resistance $R = 3 \, \Omega$.The motional electromotive force $(EMF)$ induced in the rod is given by:$\varepsilon = B l v$$\varepsilon = 0.04 \, T \times 2 \, m \times 5 \, ms^{-1} = 0.4 \, V$The current $I$ induced in the rod is given by Ohm's law:$I = \frac{\varepsilon}{R}$$I = \frac{0.4 \, V}{3 \, \Omega} = 0.1333... \, A$$I \approx 0.133 \, A = 133 \, mA$.

Class 12 Physics Electromagnetic Induction Motional EMI (Induced Parameter)

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$A$ rod of length $2 \, m$ slides with a speed of $5 \, ms^{-1}$ on a rectangular conducting frame as shown in the figure. There exists a uniform magnetic field of $0.04 \, T$ perpendicular to the plane of the figure. If the resistance of the rod is $3 \, \Omega$, the current through the rod is

KCET 2020 All KCET

A
$75 \, mA$
B
$133 \, mA$
C
$0.75 \, A$
D
$1.33 \, A$

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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 a uniform speed $v = 2 \, m/s$ 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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$A$ thin semicircular conducting ring $(PQR)$ of radius $r$ is falling with its plane vertical in a horizontal magnetic field $B,$ as shown in the figure. The potential difference developed across the ring when its speed is $v,$ is

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$A$ closed loop is moving with constant velocity in a uniform magnetic field as shown. The potential difference and current in element $AB$ will be:

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$A$ metal rod of length $1\,m$ is rotated about one of its ends in a plane at right angles to a magnetic field of induction $2.5 \times 10^{-3}\,Wb/m^2$. If it makes $1800\,rpm$,calculate the induced e.m.f. between its ends in $V$.

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$A$ rod of length $2 \, m$ slides with a speed of $5 \, ms^{-1}$ on a rectangular conducting frame as shown in the figure. There exists a uniform magnetic field of $0.04 \, T$ perpendicular to the plane of the figure. If the resistance of the rod is $3 \, \Omega$, the current through the rod is

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$A$ conducting rod $PQ$ of length $L = 1.0 \, m$ is moving with a uniform speed $v = 2 \, m/s$ 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:

$A$ thin semicircular conducting ring $(PQR)$ of radius $r$ is falling with its plane vertical in a horizontal magnetic field $B,$ as shown in the figure. The potential difference developed across the ring when its speed is $v,$ is

$A$ closed loop is moving with constant velocity in a uniform magnetic field as shown. The potential difference and current in element $AB$ will be:

$A$ conducting wire is dropped along the east-west direction. Then,

$A$ metal rod of length $1\,m$ is rotated about one of its ends in a plane at right angles to a magnetic field of induction $2.5 \times 10^{-3}\,Wb/m^2$. If it makes $1800\,rpm$,calculate the induced e.m.f. between its ends in $V$.

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