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Q.

A "L" shaped conducting rod ABC is lying in xy-plane and moving with velocity $\vec{V} = V_{0} \hat{i} + 3 V_{0} \hat{j}$ as shown in the figure. There exist a non-uniform magnetic field which depends on x and y co-ordinate as $\vec{B} = \frac{B_{0}}{L_{0}} (x + y) (- \hat{k})$ , where $B_{0}$ and $L_{0}$ are positive constants with proper dimensions. If $V_{P} - V_{Q}$ represents the motional induced emf between point P and Q. Then choose the CORRECT option(s).

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a

V_{A} - V_{B} = 18 B_{0} L_{0} V_{0}

b

V_{A} - V_{C} = \frac{27}{2} B_{0} L_{0} V_{0}

c

V_{C} - V_{A} = \frac{27}{2} B_{0} L_{0} V_{0}

d

V_{B} - V_{A} = 18 B_{0} L_{0} V_{0}

answer is A, C.

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Detailed Solution

For

V_{a} - V_{B} :

\vec{B} = \frac{B_{0}}{L_{0}} (x + L_{0}) (- \hat{k})

ε i n d = \int \vec{d l} . (\vec{V} \times \vec{B})

$V_{a} - V_{B} = \int B . d x .3 V_{0} = \frac{3 B . V_{0}}{L_{0}} \int_{L_{0}}^{3 L_{0}} (x + L_{0}) . d x$

$18 B_{0} L_{0} = V_{A} = B_{B}$

For $V_{a} - V_{B} :$ Question Image

$\vec{B} = \frac{B_{0}}{L_{0}} (3 L_{0} - y) (- \hat{k})$

$V_{C} - V_{B} = \int B . d y . V_{0} = \frac{B_{0} V_{0}}{L_{0}} \int_{L_{0}}^{2 L_{0}} (3 L_{0} + y) . d y$

$V_{C} - V_{B} = \frac{9 B_{0} L_{0}}{2}$

$V_{A} - V_{C} (18 - \frac{9}{2}) B_{0} L_{0} V_{0} = \frac{27}{2} B_{0} L_{0} V_{0}$

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