### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2003

A magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through ${60^ \circ }.$ The torque needed to maintain the needle in this position will be
A
$\sqrt 3 \,W$
B
$W$
C
${{\sqrt 3 } \over 2}W$
D
$2W$

## Explanation

$W = MB\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$

$= MB\left( {\cos {\theta ^ \circ } - \cos {{60}^ \circ }} \right)$

$= MB\left( {1 - {1 \over 2}} \right) = {{MB} \over 2}$

$\therefore$ $\tau = MB\,\sin \theta = MB\,\sin \,{60^ \circ }$

$= \sqrt 3 {{MB} \over 2} = \sqrt 3 W$
2

### AIEEE 2003

A particle of charge $- 16 \times {10^{ - 18}}$ coulomb moving with velocity $10m{s^{ - 1}}$ along the $x$-axis enters a region where a magnetic field of induction $B$ is along the $y$-axis, and an electric field of magnitude ${10^4}V/m$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis, the magnitude of $B$ is
A
${10^3}Wb/{m^2}$
B
${10^5}Wb/{m^2}$
C
${10^{16}}Wb/{m^2}$
D
${10^{ - 3}}Wb/{m^2}$

## Explanation

The situation is shown in the figure.

${F_E} =$ Force due to electric field

${F_B} =$ Force due to magnetic field

It is given that the charged particle remains moving along $X$-axis (i.e. undeviated).

Therefore ${F_B} = {F_E}$

$\Rightarrow qvB = qE$

$\Rightarrow B = {E \over v} = {{{{10}^4}} \over {10}}$

$= {10^3}\,\,weber/{m^2}$

3

### AIEEE 2003

A thin rectangular magnet suspended freely has a period of oscillation equal to $T.$ Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $T',$ the ratio ${{T'} \over T}$ is
A
${1 \over {2\sqrt 2 }}$
B
${1 \over 2}$
C
$2$
D
${1 \over 4}$

## Explanation

KEY CONCEPT : The time period of a rectangular magnet oscillating in earth's magnetic field is given by

$T = 2\pi \sqrt {{I \over {\mu {B_H}}}}$

where $I=$ Moment of inertia of the rectangular magnet

$\mu =$ Magnetic moment

${B_H} =$ Horizontal component of the earth's magnetic field

Case 1 : $T = 2\pi \sqrt {{I \over {\mu {B_H}}}}$ where $I = {1 \over {12}}M{\ell ^2}$

Case 2 : Magnet is cut into two identical pieces such that each piece has half the original length. Then

$T' = 2\pi \sqrt {{{I'} \over {\mu '{B_H}}}}$

where $I' = {1 \over {12}}\left( {{M \over 2}} \right){\left( {{\ell \over 2}} \right)^2} = {I \over 8}$ and $\mu ' = {\mu \over 2}$

$\therefore$ ${{T'} \over T} = \sqrt {{{I'} \over {\mu '}} \times {\mu \over I}}$

$= \sqrt {{{I/8} \over {\mu /2}} \times {\mu \over I}}$

$= \sqrt {{1 \over 4}} = {1 \over 2}$
4

### 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.