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A thin conducting rod of length l = 5m is moved such that its end B moves along the X-axis while end A moves along the Y-axis. An uniform magnetic field $B = 6 \hat{k} T$ exists in the region. At some instant, velocity of end B is 3 ms^-1 and the rod makes an angle of $θ = 37 °$ with the X-axis as shown in the figure. Then, at this instant, if emf induced in the rod is ' $ε$ ' then, find value of $ε / 3$ (in volts).

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answer is 7.

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Detailed Solution

$\begin{array}{l} ε = \vec{B} ({\vec{V}}_{cm} \times \vec{L}) \\ = (6 \hat{k}) ((\frac{3}{2} \hat{i} - \frac{4}{2} \hat{j}) \times (4 \hat{i} + 3 \hat{j})) \\ = 21 / 3 = 7 V . \end{array}$

Alternate method:

On considering the area under the rod

$ε = \frac{d ϕ}{d t} = B . \frac{d (x . y)}{dt} = B [x . \frac{d y}{d t} + y \frac{d x}{d t}]$

here $\frac{d y}{d t} = - 4 m / s, \frac{d x}{d t} = 3 m / s$

on substituting the values we get $ε = 21 V$

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