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The metal rod of length $l = 0.2 m$ hung horizontally on two light insulating on two light insulating wires of length L = 0.1 m in the vertical magnetic field with induction B = 1 Tesla. It is initially at $α = 60^{0}$ from the equilibrium position and released (Fig.). Find the potential difference between the ends of the rod at a time when it passes the lowest position

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$\frac{\sqrt{2}}{10} V$

0.2 V

0.1 V

Zero

answer is C.

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Detailed Solution

$m g (L - L \cos α) = \frac{1}{2} m v^{2} v^{2} = 0.1 \times 10 \Rightarrow v = 1 m / s ε = B l V$

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