A non-conducting uniform rod of length l=65m and mass m, is uniformly charged with charged q and released from rest when it is horizontal. A uniform magnetic field B0k^ is existing in space perpendicular to the plane of motion of the rod. If hinge reaction at the instant, when rod is vertical is 2mg , then the value of mgB0ql (in SI unit) is

Magnetic force does not workω=3gl N−mg+B0ωql∫0lxdx=mω2l2 N=5mg2−B023glql

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A non-conducting uniform rod of length $l = \frac{6}{5} m$ and mass m, is uniformly charged with charged q and released from rest when it is horizontal. A uniform magnetic field $B_{0} \hat{k}$ is existing in space perpendicular to the plane of motion of the rod. If hinge reaction at the instant, when rod is vertical is 2mg , then the value of $\frac{m g}{B_{0} q l}$ (in SI unit) is

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answer is 5.

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Detailed Solution

Magnetic force does not work

$ω = \sqrt{\frac{3 g}{l}} N - m g + \frac{B_{0} ω q}{l} \int_{0}^{l} x d x = \frac{m ω^{2} l}{2} N = \frac{5 m g}{2} - \frac{B_{0}}{2} \sqrt{\frac{3 g}{l}} q l$

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