### JEE Mains Previous Years Questions with Solutions

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1

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

2

### AIEEE 2007

A long straight wire of radius $a$ carries a steady current $i.$ The current is uniformly distributed across its cross section. The ratio of the magnetic field at $a/2$ and $2a$ is
A
$1/2$
B
$1/4$
C
$4$
D
$1$

## Explanation

Here, current is uniformly distributed across the cross-section of the wire, therefore, current enclosed in the ampere-an path formed at a distance ${r_1}\left( { = {a \over 2}} \right)$

$= \left( {{{\pi r_1^2} \over {\pi {a^2}}}} \right) \times I,$ where $I$ is total current

$\therefore$ Magnetic field at ${P_1}$ is

${B_1} = {{{\mu _0} \times current\,\,enclosed} \over {path}}$

$\Rightarrow {B_1} = {{{\mu _0} \times \left( {{{\pi r_1^2} \over {\pi {a^2}}}} \right) \times I} \over {2\pi {r_1}}}$

$= {{{\mu _0} \times I{r_1}} \over {2\pi {a^2}}}$

Now, magnetic field at point ${P_2},$

${B_2} = {{{\mu _0}} \over {2\pi }}.{I \over {\left( {2a} \right)}} = {{{\mu _0}I} \over {4\pi a}}$

$\therefore$ Required ratio $= {{{B_1}} \over {{B_2}}} = {{{\mu _0}I{r_1}} \over {2\pi {a^2}}} \times {{4\pi a} \over {{\mu _0}I}}$

$= {{2{r_1}} \over a} = {{2 \times {a \over 2}} \over a} = 1.$
3

### AIEEE 2007

A current $I$ flows along the length of an infinitely long, straight, thin walled pipe. Then
A
the magnetic field at all points inside the pipe is the same, but not zero
B
the magnetic field is zero only on the axis of the pipe
C
the magnetic field is different at different points inside the pipe
D
the magnetic field at any point inside the pipe is zero

## Explanation

There is no current inside the pipe. Therefore

$\oint {\overline B .\overline {d\ell } } = {\mu _0}I$

$I=0$

$\therefore$ $B=0$
4

### AIEEE 2007

A charged particle moves through a magnetic field perpendicular to its direction. Then
A
Kinetic energy changes but the momentum is constant
B
the momentum changes but the kinetic energy is constant
C
both momentum and kinetic energy of the particle are not constant
D
both momentum and kinetic energy of the particle are constant

## Explanation

NOTE : When a charged particle enters a magnetic field at a direction perpendicular to the direction of motion, the path of the motion is circular. In circular motion the direction of velocity changes at every point (the magnitude remains constant).

Therefore, the tangential momentum will change at every point. But kinetic energy will remain constant as it is given by ${1 \over 2}\,m{v^2}$ and ${v^2}$ is the square of the magnitude of velocity which does not change.