A metallic cube of side 15,cm is moving along | vedclass

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Question

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

Vedclass · Accepted Answer

$150$

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(A) The motional electromotive force $(EMF)$ induced across a conductor moving in a magnetic field is given by the formula $\varepsilon = (\vec{v} 	imes \vec{B}) \cdot \vec{l}$.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 $\vec{v} 	imes \vec{B} = (2\hat{j}) 	imes (0.5\hat{k}) = 1\hat{i}\,V/m$.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 $\Delta V = |\vec{v} 	imes \vec{B}| 	imes l = 1\,V/m 	imes 0.15\,m = 0.15\,V$.Converting to millivolts,$\Delta V = 0.15 	imes 1000\,mV = 150\,mV$.

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