A horizontal straight conductor of mass ‘m’ and length ‘l’ is placed in a uniform vertical magnetic field of magnitude ‘B’. An amount of charge ‘Q’ passes through the rod in a very short time such that the conductor begins to move only after all the charge has passed through it. Its initial velocity will be

$\frac{B Q l}{m}$

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$\frac{B Q l}{2 m}$

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$\frac{3 B Q l}{2 m}$

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The conductor does not move

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Solution

The correct option is A
$\frac{B Q l}{m}$

$m v - m u = \int F d t o r u = 0 & F = i l B = \frac{d q}{d t} l B \Rightarrow m v = \int_{0}^{q} \frac{d q}{d t} \Rightarrow v = \frac{Q l B}{m}$

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The correct option is A BQlm mv−mu=∫F dt or u=0 & F=ilB=dqdtlB⇒mv=∫q0dqdt⇒v=QlBm

The correct option is A
$\frac{B Q l}{m}$

$m v - m u = \int F d t o r u = 0 & F = i l B = \frac{d q}{d t} l B \Rightarrow m v = \int_{0}^{q} \frac{d q}{d t} \Rightarrow v = \frac{Q l B}{m}$