JEE Mains Previous Years Questions with Solutions

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1

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

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

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

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

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.
3

AIEEE 2002

If an electron and a proton having same momentum enter perpendicular to a magnetic field, then
A
curved path of electron and proton will be same (ignoring the sense of revolution)
B
they will move undeflected
C
curved path of electron is more curved than that of the proton
D
path of proton is more curved.

Explanation

KEY CONCEPT : When a charged particle enters perpendicular to a magnetic field,

then it moves in a circular path of radius.

$r = {p \over {qB}}$

where $q=$ Charge of the particle

$p=$ Momentum of the particle

$B=$ Magnetic field

Here $p,q$ and $B$ are constant for electron and proton, therefore the radius will be

same.
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

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

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