A conducting rod PQ of length 5,m oriented as | vedclass

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(A) The induced $EMF$ in a moving conductor is given by the formula $\varepsilon = (\vec{v} 	imes \vec{B}) \cdot \vec{L}$.Given:Velocity $\vec{v} = 2

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$32$

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(A) The induced $EMF$ in a moving conductor is given by the formula $\varepsilon = (\vec{v} 	imes \vec{B}) \cdot \vec{L}$.Given:Velocity $\vec{v} = 2\hat{i}\,m/s$Magnetic field $\vec{B} = 3\hat{j} + 4\hat{k}\,T$Length vector $\vec{L} = 5\cos(53^\circ)\hat{i} + 5\sin(53^\circ)\hat{j} = 5(0.6)\hat{i} + 5(0.8)\hat{j} = (3\hat{i} + 4\hat{j})\,m$.Now,calculate the cross product $\vec{v} 	imes \vec{B}$:$\vec{v} 	imes \vec{B} = (2\hat{i}) 	imes (3\hat{j} + 4\hat{k}) = 6(\hat{i} 	imes \hat{j}) + 8(\hat{i} 	imes \hat{k}) = 6\hat{k} - 8\hat{j}$.Now,calculate the dot product with $\vec{L}$:$\varepsilon = (6\hat{k} - 8\hat{j}) \cdot (3\hat{i} + 4\hat{j}) = (6\hat{k} \cdot 3\hat{i}) + (6\hat{k} \cdot 4\hat{j}) - (8\hat{j} \cdot 3\hat{i}) - (8\hat{j} \cdot 4\hat{j}) = 0 + 0 - 0 - 32 = -32\,V$.The magnitude of the induced $EMF$ is $32\,V$.

$A$ conducting rod $PQ$ of length $5\,m$ oriented as shown in the figure is moving with a velocity $\vec{v} = (2\,m/s)\hat{i}$ without any rotation in a uniform magnetic field $\vec{B} = (3\hat{j} + 4\hat{k})\,T$. The induced $EMF$ in the rod is........$V$.

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