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(A) The induced emf is given by $\varepsilon = (\vec{v} 	imes \vec{B}) \cdot \vec{L}$.Since the wire is perpendicular to the $xy$-plane,its length ve

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$2 \ V$

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(A) The induced emf is given by $\varepsilon = (\vec{v} 	imes \vec{B}) \cdot \vec{L}$.Since the wire is perpendicular to the $xy$-plane,its length vector is $\vec{L} = 2\hat{k} \ m$.The velocity is $\vec{v} = (2\hat{i} + 3\hat{j} + \hat{k}) \ m/s$ and the magnetic field is $\vec{B} = (\hat{i} + 2\hat{j}) \ Wb/m^2$.First,calculate the cross product $\vec{v} 	imes \vec{B}$:$\vec{v} 	imes \vec{B} = (2\hat{i} + 3\hat{j} + \hat{k}) 	imes (\hat{i} + 2\hat{j})$$= 2(\hat{i} 	imes \hat{i}) + 4(\hat{i} 	imes \hat{j}) + 3(\hat{j} 	imes \hat{i}) + 6(\hat{j} 	imes \hat{j}) + 1(\hat{k} 	imes \hat{i}) + 2(\hat{k} 	imes \hat{j})$$= 0 + 4\hat{k} - 3\hat{k} + 0 + \hat{j} - 2\hat{i} = -2\hat{i} + \hat{j} + \hat{k}$.Now,calculate the dot product with $\vec{L}$:$\varepsilon = (-2\hat{i} + \hat{j} + \hat{k}) \cdot (2\hat{k})$$= (-2 	imes 0) + (1 	imes 0) + (1 	imes 2) = 2 \ V$.

$A$ thin wire of length $2 \ m$ is perpendicular to the $xy$-plane. It moves with a velocity $v = (2\hat{i} + 3\hat{j} + \hat{k}) \ m/s$ in a magnetic field $B = (\hat{i} + 2\hat{j}) \ Wb/m^2$. What is the induced potential difference (emf) across the ends of the wire?

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