### JEE Mains Previous Years Questions with Solutions

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1

### 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.
2

### 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.
3

### 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.

The magnitude of the magnetic field $(B)$ due to the loop $ABCD$ at the origin $(O)$ is :

A
${{{\mu _0}I\left( {b - a} \right)} \over {24ab}}$
B
${{{\mu _0}I} \over {4\pi }}\left[ {{{b - a} \over {ab}}} \right]$
C
${{{\mu _0}I} \over {4\pi }}\left[ {2\left( {b - a} \right) + {\raise0.5ex\hbox{\scriptstyle \pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 3}}\left( {a + b} \right)} \right]$
D
zero

## Explanation

The magnetic field at $O$ due to current in $DA$ is

${B_1} = {{{\mu _0}} \over {4\pi }}{I \over a} \times {\pi \over 6}\,\,\,$ (directed vertically upwards)

The magnetic field at $O$ due to current in $BC$ is

${B_2} = {{{\mu _0}} \over {4\pi }}{I \over b} \times {\pi \over 6}$ (directed vertically downwards)

The magnetic field due to current $AB$ and $CD$ at $O$ is zero.

therefore the net magnetic field is

$B = {B_1} - {B_2}$ (directed vertically upwards)

$= {{{\mu _0}} \over {4\pi }}{I \over a}{\pi \over 6} - {{{\mu _0}} \over {4\pi }}{I \over b} \times {\pi \over 6}$

$= {{{\mu _0}I} \over {24}}\left( {{1 \over a} - {1 \over b}} \right)$

$= {{{\mu _0}I} \over {24ab}}\left( {b - a} \right)$
4

### AIEEE 2008

A horizontal overhead powerline is at height of $4m$ from the ground and carries a current of $100A$ from east to west. The magnetic field directly below it on the ground is
$\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,\,Tm\,\,{A^{ - 1}}} \right)$
A
$2.5 \times {10^{ - 7}}\,T$ southward
B
$5 \times {10^{ - 6}}\,T$ northward
C
$5 \times {10^{ - 6}}\,T$ southward
D
$2.5 \times {10^{ - 7}}\,T$ northward

## Explanation

The magnetic field is

$B = {{{\mu _0}} \over {4\pi }}{{2I} \over r}$

$= {10^{ - 7}} \times {{2 \times 100} \over 4}$

$= 5 \times {10^{ - 6}}T$

According to right hand palm rule, the magnetic field is directed towards south.