### JEE Mains Previous Years Questions with Solutions

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### 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, ${r_\alpha } = {r_p} < {r_d}$
2

### 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,

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

$\int\limits_0^\pi {\sin \theta d\theta = {{{\mu _0}I} \over {{\pi ^2}R}}}$
3

### AIEEE 2010

Two long parallel wires are at a distance $2d$ apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field $B$ along the line $XX'$ is given by
A
B
C
D

## Explanation

The magnetic field varies inversely with the distance for a long conductor. That is, $B \propto {1 \over d}.$ According to the magnitude and direction shown graph $(1)$ is the correct one.
4

### AIEEE 2009

A current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $BC$ (radius $= b$) and $DA$ (radius $=a$) of the loop are joined by two straight wires $AB$ and $CD$. A steady current $I$ is flowing in the loop. Angle made by $AB$ and $CD$ at the origin $O$ is ${30^ \circ }.$ Another straight thin wire steady current ${I_1}$ flowing out of the plane of the paper is kept at the origin.

Due to the presence of the current ${I_1}$ at the origin:

A
The forces on $AD$ are $BC$ are zero.
B
The magnitude of the net force on the loop is given by ${{{I_1}I} \over {4\pi }}{\mu _0}\left[ {2\left( {b - a} \right) + {\raise0.5ex\hbox{\scriptstyle \pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 3}}\left( {a + b} \right)} \right].$
C
The magnitude of the net force on the loop is given by ${{{\mu _0}I{I_1}} \over {24ab}}\left( {b - a} \right).$
D
The forces on $AB$ and $DC$ are zero.

## Explanation

KEY CONCEPT : $\overrightarrow F = I\left( {\overrightarrow \ell \times \overrightarrow B } \right)$

The force on $AD$ and $BC$ due to current ${I_1}$ is zero. This is because the directions of current element $I\overrightarrow {d\ell }$ and magnetic field $\overrightarrow B$ are parallel.