The conductor of length \(l\) is placed perpendicular to a horizontal uniform magnetic field \(B\). Suddenly a certain amount of charge is passed through it, when it is found to jump to a height \(h\). The amount of charge that passes through the conductor is

\((a)\) \(\frac{m \sqrt{gh}}{Bl}\)

\((b)\) \(\frac{m \sqrt{gh}}{2Bl}\)

\((c)\) \(\frac{m \sqrt{2gh}}{Bl}\)

\((d)\) None of these

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TopNewestUsing conservation of energy, \(\Delta\)K=\(\Delta\)U,

or m\(v^{2}\)/2=mgh -(i)

acceleration, a=

\(F_{m}\)/m=ilB/m=qlB/mt -(ii)Now use v=u + at, where u=0, a is got from (ii) and v is obtained from (i) to solve for q.

I think (c) is correct. Please confirm from others to verify. BTW what's your score in JEE-MAINS ? – Nishant Sharma · 3 years, 8 months ago

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– Advitiya Brijesh · 3 years, 8 months ago

yeah!!! thanks a lotLog in to reply