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(B) The induced electromotive force (e.m.f.) in a moving conductor is given by the formula $\varepsilon = (\vec{v} 	imes \vec{B}) \cdot \vec{L}$.Give

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$25$

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(B) The induced electromotive force (e.m.f.) in a moving conductor is given by the formula $\varepsilon = (\vec{v} 	imes \vec{B}) \cdot \vec{L}$.Given:Velocity $\vec{v} = 1\hat{i} \, ms^{-1}$.Magnetic field $\vec{B} = 3\hat{i} + 4\hat{j} + 5\hat{k} \, T$.Length vector $\vec{L} = 5\hat{j} \, m$ (since it is placed along the $y$-axis).First,calculate the cross product $(\vec{v} 	imes \vec{B})$:$\vec{v} 	imes \vec{B} = (1\hat{i}) 	imes (3\hat{i} + 4\hat{j} + 5\hat{k}) = 4\hat{k} - 5\hat{j}$.Now,calculate the dot product with $\vec{L}$:$\varepsilon = (4\hat{k} - 5\hat{j}) \cdot (5\hat{j}) = -25 \, V$.The magnitude of the induced e.m.f. is $|\varepsilon| = 25 \, V$.

$A$ uniform magnetic field exists in the region given by $\vec{B} = 3\hat{i} + 4\hat{j} + 5\hat{k} \, T$. $A$ rod of length $L = 5 \, m$ placed along the $y$-axis is moved along the $x$-axis with a constant speed $v = 1 \, ms^{-1}$. The induced e.m.f. in the rod is ......... $V$.

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