(A) The motional electromotive force $(EMF)$ induced across a conductor moving in a magnetic field is given by the formula $\varepsilon = (\vec{v} \ti

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(A) The motional electromotive force $(EMF)$ induced across a conductor moving in a magnetic field is given by the formula $\varepsilon = (\vec{v} \times \vec{B}) \cdot \vec{l}$.Here,the velocity vector is $\vec{v} = 2\hat{j}\,m/s$ and the magnetic field vector is $\vec{B} = 0.5\hat{k}\,T$.The cross product is $\vec{v} \times \vec{B} = (2\hat{j}) \times (0.5\hat{k}) = 1\hat{i}\,V/m$.This indicates that the induced electric field is along the $x$-axis.The length of the cube along the $x$-axis is $l = 15\,cm = 0.15\,m$.The potential difference is $\Delta V = |\vec{v} \times \vec{B}| \times l = 1\,V/m \times 0.15\,m = 0.15\,V$.Converting to millivolts,$\Delta V = 0.15 \times 1000\,mV = 150\,mV$.

Class 12 Physics Electromagnetic Induction Motional EMI (Induced Parameter)

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$A$ metallic cube of side $15\,cm$ is moving along the $y$-axis at a uniform velocity of $2\,m/s$ in a region of uniform magnetic field of magnitude $0.5\,T$ directed along the $z$-axis. In equilibrium,the potential difference between the faces of higher and lower potential developed because of the motion through the field will be $..........mV$.

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Motional EMI (Induced Parameter)

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$A$ metallic rod of length $1 \,m$ held along the east-west direction is allowed to fall down freely. Given the horizontal component of the Earth's magnetic field $B_H = 3 \times 10^{-5} \,T$. The emf induced in the rod at an instant $t = 2 \,s$ after it is released is (Take $g = 10 \,m/s^2$):

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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 with radial metal spokes $1 \ m$ in length is rotated in a magnetic field of $0.5 \times 10^{-4} \ T$ normal to the plane of the wheel. If the induced emf between the rim and axle is $\pi / 3000 \ V$,then the rotational speed of the wheel in revolutions per minute is

MediumTS EAMCET 2018

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$A$ rectangular wire loop of sides $8 \text{ cm}$ and $3 \text{ cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3 \text{ T}$ directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is $2 \text{ cm s}^{-1}$ in a direction normal to the shorter side of the loop, will be:

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$A$ rod of length $l$ rotates with a uniform angular velocity $\omega$ about an axis passing through its middle point but normal to its length in a uniform magnetic field of induction $B$ with its direction parallel to the axis of rotation. The induced $emf$ between the two ends of the rod is

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$A$ metallic rod of length $1 \,m$ held along the east-west direction is allowed to fall down freely. Given the horizontal component of the Earth's magnetic field $B_H = 3 \times 10^{-5} \,T$. The emf induced in the rod at an instant $t = 2 \,s$ after it is released is (Take $g = 10 \,m/s^2$):

$A$ 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 . . . . . . .

$A$ wheel with radial metal spokes $1 \ m$ in length is rotated in a magnetic field of $0.5 \times 10^{-4} \ T$ normal to the plane of the wheel. If the induced emf between the rim and axle is $\pi / 3000 \ V$,then the rotational speed of the wheel in revolutions per minute is

$A$ rod of length $l$ rotates with a uniform angular velocity $\omega$ about an axis passing through its middle point but normal to its length in a uniform magnetic field of induction $B$ with its direction parallel to the axis of rotation. The induced $emf$ between the two ends of the rod is

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