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Question

(C) Let the magnetic field be $\vec{B} = B \hat{k}$.The motional electromotive force $(EMF)$ induced in a conductor moving in a magnetic field is give

Vedclass · Accepted Answer

$v_x l_2 - v_y l_1$

Vedclass · Answer

(C) Let the magnetic field be $\vec{B} = B \hat{k}$.The motional electromotive force $(EMF)$ induced in a conductor moving in a magnetic field is given by $\int (\vec{v} 	imes \vec{B}) \cdot d\vec{l}$.For segment $AB$,the length vector is $\vec{l}_{AB} = l_1 \hat{i}$.The potential difference $V_A - V_B = \int_A^B (\vec{v} 	imes \vec{B}) \cdot d\vec{l} = (v_x \hat{i} + v_y \hat{j}) 	imes (B \hat{k}) \cdot (l_1 \hat{i}) = (-v_x B \hat{j} + v_y B \hat{i}) \cdot (l_1 \hat{i}) = v_y B l_1$.So,$V_A - V_B = B v_y l_1$.For segment $BC$,the length vector is $\vec{l}_{BC} = l_2 \hat{j}$.The potential difference $V_B - V_C = \int_B^C (\vec{v} 	imes \vec{B}) \cdot d\vec{l} = (v_x \hat{i} + v_y \hat{j}) 	imes (B \hat{k}) \cdot (l_2 \hat{j}) = (-v_x B \hat{j} + v_y B \hat{i}) \cdot (l_2 \hat{j}) = -v_x B l_2$.So,$V_B - V_C = -B v_x l_2$.Adding the two potential differences:$(V_A - V_B) + (V_B - V_C) = B v_y l_1 - B v_x l_2$.$V_A - V_C = B(v_y l_1 - v_x l_2)$.The magnitude of the potential difference $|V_A - V_C| = B |v_x l_2 - v_y l_1|$.Thus,the potential difference is proportional to $v_x l_2 - v_y l_1$.

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