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

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(A) The length vector of the rod is $\vec{l} = 5 \cos 53^{\circ} \hat{i} + 5 \sin 53^{\circ} \hat{j} = 3 \hat{i} + 4 \hat{j}$.The induced $Emf$ $(e)$

Vedclass · Accepted Answer

$32$

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(A) The length vector of the rod is $\vec{l} = 5 \cos 53^{\circ} \hat{i} + 5 \sin 53^{\circ} \hat{j} = 3 \hat{i} + 4 \hat{j}$.The induced $Emf$ $(e)$ in a moving conductor is given by the formula $e = |(\vec{v} 	imes \vec{B}) \cdot \vec{l}|$.Given $\vec{v} = 2\hat{i}$,$\vec{B} = 3\hat{j} + 4\hat{k}$,and $\vec{l} = 3\hat{i} + 4\hat{j}$.First,calculate the cross product $\vec{v} 	imes \vec{B} = (2\hat{i}) 	imes (3\hat{j} + 4\hat{k}) = 6(\hat{i} 	imes \hat{j}) + 8(\hat{i} 	imes \hat{k}) = 6\hat{k} - 8\hat{j}$.Now,calculate the dot product with $\vec{l}$:$e = |(6\hat{k} - 8\hat{j}) \cdot (3\hat{i} + 4\hat{j})| = |(6\hat{k} \cdot 3\hat{i}) + (6\hat{k} \cdot 4\hat{j}) + (-8\hat{j} \cdot 3\hat{i}) + (-8\hat{j} \cdot 4\hat{j})|$.Since $\hat{i} \cdot \hat{j} = 0$,$\hat{j} \cdot \hat{k} = 0$,and $\hat{i} \cdot \hat{k} = 0$,and $\hat{j} \cdot \hat{j} = 1$:$e = |0 + 0 + 0 - 32(1)| = |-32| = 32\,V$.

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

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