### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2013 (Offline)

Two short bar magnets of length $1$ $cm$ each have magnetic moments $1.20$ $A{m^2}$ and $1.00$ $A{m^2}$ respectively. They are placed on a horizontal table parallel to each other with their $N$ poles pointing towards the South. They have a common magnetic equator and are separated by a distance of $20.0$ $cm.$ The value of the resultant horizontal magnetic induction at the mid-point $O$ of the line joining their centres is close to $\left( \, \right.$ Horizontal com-ponent of earth. $s$ magnetic induction is $3.6 \times 10.5Wb/{m^2})$
A
$3.6 \times 10.5\,\,Wb/{m^2}$
B
$2.56 \times 10.4\,\,Wb/{m^2}$
C
$3.50 \times 10.4\,\,Wb/{m^2}$
D
$5.80 \times 10.4\,Wb/{m^2}$

## Explanation

Given: ${M_1} = 1.20A{m^2}\,\,\,$ and $\,\,\,{M_2} = 1.00A{m^2}$

$r = {{20} \over 2}cm = 0.1m$

${B_{net}} = {B_1} + {B_2} + {B_H}$

${B_{net}} = {{{\mu _0}\left( {{M_1} + {M_2}} \right)} \over {{r^3}}} + {B_H}$

$= {{{{10}^{ - 7}}\left( {1.2 + 1} \right)} \over {{{\left( {0.1} \right)}^3}}} + 3.6 \times {10^{ - 5}}$

$= 2.56 \times {10^{ - 4}}\,\,wb/{m^2}$
2

### AIEEE 2012

A charge $Q$ is uniformly distributed over the surface of non-conducting disc of radius $R.$ The disc rotates about an axis perpendicular to its plane and passing through its center with an angular velocity $\omega .$ As a result of this rotation a magnetic field of induction $B$ is obtained at the center of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and very the radius of the disc then the variation of the magnetic induction at the center of the disc will be represented by the figure :
A
B
C
D

## Explanation

The magnetic field due a disc is given as

$B = {{{h_0}\omega Q} \over {2\pi R}}$ i.e., $B \propto {1 \over R}$
3

### AIEEE 2012

Proton, deuteron and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, denuteron and alpha particle are respectively ${r_p},{r_d}$ and ${r_\alpha }$. Which one of the following relation is correct?
A
${r_\alpha } = {r_p} = {r_d}$
B
${r_\alpha } = {r_p} < {r_d}$
C
${r_\alpha } > {r_d} > {r_p}$
D
${r_\alpha } = {r_d} > {r_p}$

## Explanation

$r = {{\sqrt {2mv} } \over {qB}} \Rightarrow r \times v{{\sqrt m } \over q}$

Thus we have, ${r_\alpha } = {r_p} < {r_d}$
4

### AIEEE 2011

A current $I$ flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius $R.$ The magnitude of the magnetic induction along its axis is:
A
${{{\mu _0}I} \over {2{\pi ^2}R}}$
B
${{{\mu _0}I} \over {2\pi R}}$
C
${{{\mu _0}I} \over {4\pi R}}$
D
${{{\mu _0}I} \over {{\pi ^2}R}}$

## Explanation

Current in a small element, $dl = {{d\theta } \over \pi }I$

Magnetic field due to the element

$dB = {{{\mu _0}} \over {4\pi }}{{2dl} \over R}$

The component $dB$ $\cos \,\theta ,$ of the field is canceled by another opposite component.

Therefore,

${B_{net}} = \int {dB\sin \theta = {{{\mu _0}I} \over {2{\pi ^2}{R_0}}}}$

$\int\limits_0^\pi {\sin \theta d\theta = {{{\mu _0}I} \over {{\pi ^2}R}}}$