### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### 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)

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

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

$= {{{\mu _0}I} \over {24ab}}\left( {b - a} \right)$
2

### 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.
3

### AIEEE 2008

Relative permittivity and permeability of a material ${\varepsilon _r}$ and ${\mu _r},$ respectively. Which of the following values of these quantities are allowed for a diamagnetic material?
A
${\varepsilon _r} = 0.5,\,\,{\mu _r} = 1.5$
B
${\varepsilon _r} = 1.5,\,\,{\mu _r} = 0.5$
C
${\varepsilon _r} = 0.5,\,\,{\mu _r} = 0.5$
D
${\varepsilon _r} = 1.5,\,\,{\mu _r} = 1.5$

## Explanation

For a diamagnetic material, the value of ${\mu _r}$ is less than one. For any material, the value of ${ \in _r}$ is always greater than $1.$
4

### AIEEE 2007

Two identical conducting wires $AOB$ and $COD$ are placed at right angles to each other. The wire $AOB$ carries an electric current ${I_1}$ and $COD$ carries a current ${I_2}$. The magnetic field on a point lying at a distance $d$ from $O$, in a direction perpendicular to the plane of the wires $AOB$ and $COD$ , will be given by
A
${{{\mu _0}} \over {2\pi d}}\left( {I_1^2 + I_2^2} \right)$
B
${{{\mu _0}} \over {2\pi }}{\left( {{{{I_1} + {I_2}} \over d}} \right)^{{1 \over 2}}}$
C
${{{\mu _0}} \over {2\pi d}}{\left( {I_1^2 + I_2^2} \right)^{{1 \over 2}}}$
D
${{{\mu _0}} \over {2\pi d}}\left( {{I_1} + {I_2}} \right)$

## Explanation

Clearly, the magnetic fields at a point $P,$ equidistant from $AOB$ and $COD$ will have directions perpendicular to each other, as they are placed normal to each other.

$\therefore$ Resultant field, $B = \sqrt {B_1^2 + B_2^2}$

But ${B_1} = {{{\mu _0}{I_1}} \over {2\pi d}}$ and ${B_2} = {{{\mu _0}{I_2}} \over {2\pi d}}$

$\therefore$ $B = \sqrt {{{\left( {{{{\mu _0}} \over {2\pi d}}} \right)}^2}\left( {I_1^2 + I_2^2} \right)}$

or, $B = {{{\mu _0}} \over {2\pi d}}{\left( {I_1^2 + I_2^2} \right)^{1/2}}$