A conductor of length l and mass m is placed along the east-west line on a table. Suddenly a certain amount of change is passed through it and it is found to jump to a height h. The earth's magnetic induction is B. The charge passed through the conductor is:

\frac{1}{B m g h}

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\frac{\sqrt{2 g h}}{B / m}

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\frac{g h}{B / m}

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\frac{m \sqrt{2 g h}}{B l}

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Solution

The correct option is B $\frac{\sqrt{2 g h}}{B / m}$
Now
$\begin{matrix} F o r c e p r o d u c e d i n t h e r o d : F = i l B F = \frac{d i}{d t} l b I f a c c e l e r a t i o n b e a, M a = \frac{Δ Q}{Δ T} l B M Δ T = Q l B M V = Q l B S i n c e e n e r g y c o n s e r v a t i o n \frac{1}{2} m v^{2} = n g h v = \sqrt{2 g h} Q i B = \sqrt{2 g h} Q = \frac{\sqrt{2 g h}}{B l} H e n c e, O P t i o n B i s t h e c o r r e c t a n s w e r . \end{matrix}$

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A conductor of length l and mass m is placed along the east-west line on a table. Suddenly a certain amount of change is passed through it and it is found to jump to a height h. The earth's magnetic induction is B. The charge passed through the conductor is:

The correct option is B √2ghB/mNowForceproducedintherod:F=ilBF=didtlbIfaccelerationbea,Ma=ΔQΔTlBMΔT=QlBMV=QlBSinceenergyconservation12mv2=nghv=√2ghQiB=√2ghQ=√2ghBlHence,OPtionBisthecorrectanswer.