A non-conducting uniform rod of length l=65m and mass m, is uniformly charged with charge 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 g=10 m/s2

Magnetic force does not do work.ω=3gl N−mg+B0ωql∫0lxdx=mω2l2 N=5mg2−B023gl qlSubstituting, N=2mg, we get, mgB0ql=3gl=3×10×56=5

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A non-conducting uniform rod of length $l = \frac{6}{5} m$ and mass $m$ , is uniformly charged with charge $q$ 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 $2 m g$ , then the value of $\frac{m g}{B_{0} q l}$ (in SI unit) is $(g = 10 m / s^{2})$

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answer is 5.

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Detailed Solution

Magnetic force does not do work.

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

Substituting, $N = 2 m g$ , we get, $\frac{m g}{B_{0} q l} = \sqrt{\frac{3 g}{l}} = \sqrt{\frac{3 \times 10 \times 5}{6}} = 5$

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