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

Easy Math Editor

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Comments

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 ?

Nishant Sharma - 8 years ago

yeah!!! thanks a lot

Advitiya Brijesh - 8 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$