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

5 years, 1 month ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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 ?

- 5 years, 1 month ago

yeah!!! thanks a lot

- 5 years, 1 month ago