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# Magnetics!!!

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

4 years, 8 months ago

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Using 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 ?

- 4 years, 8 months ago

yeah!!! thanks a lot

- 4 years, 8 months ago