### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2003

A particle of mass $M$ and charge $Q$ moving with velocity $\overrightarrow v$ describe a circular path of radius $R$ when subjected to a uniform transverse magnetic field of induction $B.$ The network done by the field when the particle completes one full circle is
A
$\left( {{{M{v^2}} \over R}} \right)2\pi R$
B
zero
C
$B\,\,Q\,2\pi R$
D
$B\,Qv\,2\pi R$

## Explanation

The work done, $dW = Fds\,\cos \,\theta$

The angle between force and displacement is ${90^ \circ }.$

Therefore work done is zero.
2

### AIEEE 2002

Wires $1$ and $2$ carrying currents $i{}_1$ and $i{}_2$ respectively are inclined at an angle $\theta$ to each other. What is the force on a small element $dl$ of wire $2$ at a distance of $r$ from wire $1$ (as shown in figure) due to the magnetic field of wire $1$?
A
${{{\mu _0}} \over {2\pi r}}{i_1}{i_2}\,dl\,\tan \,\theta$
B
${{{\mu _0}} \over {2\pi r}}{i_1}{i_2}\,dl\,\sin \,\theta$
C
${{{\mu _0}} \over {2\pi r}}{i_1}{i_2}\,dl\,\cos \,\theta$
D
${{{\mu _0}} \over {4\pi r}}{i_1}{i_2}\,dl\,\sin \,\theta$

## Explanation

Magnetic field due to current in wire $1$ at point $P$ distant $r$ from the wire is

$B = {{{\mu _0}} \over {4\pi }}{{{i_1}} \over r}\left[ {\cos \theta + \cos \theta } \right]$

$B = {{{\mu _0}} \over {2\pi }}{{{i_1}\cos \theta } \over r}$ (directed perpendicular to the plane of paper, inwards)

The force exerted due to this magnetic field on current element ${i_2}\,dl$ is

$dF = {i_2}\,dl\,B\,\sin \,{90^ \circ }$

$\therefore$ $dF = {i_2}dl\left[ {{{{\mu _0}} \over {2\pi }}{{{i_1}\cos \theta } \over r}} \right]$

$= {{{\mu _0}} \over {2\pi r}}{i_1}\,{i_2}dl\,\cos \,\theta$
3

### AIEEE 2002

The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its
A
speed
B
mass
C
charge
D
magnetic induction

## Explanation

KEY CONCEPT : The time period of a charged particle

$\left( {m,q} \right)$ moving in a magnetic field $(B)$ is $T = {{2\pi m} \over {qB}}$

The time period does not depend on the speed of the particle.
4

### AIEEE 2002

If in a circular coil $A$ of radius $R,$ current $I$ is flowing and in another coil $B$ of radius $2R$ a current $2I$ is flowing, then the ratio of the magnetic fields ${B_A}$ and ${B_B}$, produced by them will be
A
$1$
B
$2$
C
$1/2$
D
$4$

## Explanation

KEY CONCEPT : We know that the magnetic field produced by a current carrying circular coil of radius $r$

at its center is $B = {{{\mu _0}} \over {4\pi }}{I \over r} \times 2\pi$

Here ${B_A} = {{{\mu _0}} \over {4\pi }}{I \over R} \times 2\pi$ and

${B_B} = {{{\mu _0}} \over {4\pi }}{{2I} \over {2R}} \times 2\pi$

$\Rightarrow {{{B_A}} \over {B{}_B}} = 1$

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