### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2014 (Offline)

A conductor lies along the $z$-axis at $- 1.5 \le z < 1.5\,m$ and carries a fixed current of $10.0$ $A$ in $- {\widehat a_z}$ direction (see figure). For a field $\overrightarrow B = 3.0 \times {10^{ - 4}}\,{e^{ - 0.2x}}\,\,{\widehat a_y}\,\,T,$ find the power required to move the conductor at constant speed to $x=2.0$ $m$, $y=0$ $m$ in $5 \times {10^{ - 3}}s.$ Assume parallel motion along the $x$-axis.
A
$1.57W$
B
$2.97W$
C
$14.85$ $W$
D
$29.7W$

## Explanation

Work done in moving the conductor is,

$W = \int_0^2 {Fdx}$

$= \int_0^2 {3.0 \times {{10}^{ - 4}}\,{e^{ - 02x}} \times 10 \times 3dx}$

$= 9 \times {10^{ - 3}}\int_0^2 {{e^{ - 0.2x}}\,dx}$

$= {{9 \times {{10}^{ - 3}}} \over {0.2}}\left[ { - {e^{ - 0.2 \times 2}} + 1} \right]\,\,\,$

$B = 3.0 \times {10^{ - 4}}\,{e^{ - 0.2x}}$

(By exponential function)

$= {{9 \times {{10}^{ - 3}}} \over {0.2}} \times \left[ {1 - {e^{ - 0.4}}} \right]$

$= 9 \times {10^{ - 3}} \times \left( {0.33} \right) = 2.97 \times {10^{ - 3}}J$

Power required to move the conductor is,

$P = {W \over t}$

$P = {{2.97 \times {{10}^{ - 3}}} \over {\left( {0.2} \right) \times 5 \times {{10}^{ - 3}}}} = 2.97W$
2

### JEE Main 2014 (Offline)

The coercivity of a small magnet where the ferromagnet gets demagnetized is $3 \times {10^3}\,A{m^{ - 1}}.$ The current required to be passed in a solenoid of length $10$ $cm$ and number of turns $100,$ so that the magnet gets demagnetized when inside the solenoid, is :
A
$30$ $mA$
B
$60$ $mA$
C
$3$ $A$
D
$6A$

## Explanation

Magnetic field in solenoid $B = {\mu _0}ni$

$\Rightarrow {B \over {{\mu _0}}} = ni$

(Where $n=$ number of turns per unit length)

$\Rightarrow {B \over {{\mu _0}}} = {{Ni} \over L}$

$\Rightarrow 3 \times {10^3} = {{100i} \over {10 \times {{10}^{ - 2}}}}$

$\Rightarrow i = 3A$
3

### JEE Main 2013 (Offline)

Two short bar magnets of length $1$ $cm$ each have magnetic moments $1.20$ $A{m^2}$ and $1.00$ $A{m^2}$ respectively. They are placed on a horizontal table parallel to each other with their $N$ poles pointing towards the South. They have a common magnetic equator and are separated by a distance of $20.0$ $cm.$ The value of the resultant horizontal magnetic induction at the mid-point $O$ of the line joining their centres is close to $\left( \, \right.$ Horizontal com-ponent of earth. $s$ magnetic induction is $3.6 \times 10.5Wb/{m^2})$
A
$3.6 \times 10.5\,\,Wb/{m^2}$
B
$2.56 \times 10.4\,\,Wb/{m^2}$
C
$3.50 \times 10.4\,\,Wb/{m^2}$
D
$5.80 \times 10.4\,Wb/{m^2}$

## Explanation

Given: ${M_1} = 1.20A{m^2}\,\,\,$ and $\,\,\,{M_2} = 1.00A{m^2}$

$r = {{20} \over 2}cm = 0.1m$

${B_{net}} = {B_1} + {B_2} + {B_H}$

${B_{net}} = {{{\mu _0}\left( {{M_1} + {M_2}} \right)} \over {{r^3}}} + {B_H}$

$= {{{{10}^{ - 7}}\left( {1.2 + 1} \right)} \over {{{\left( {0.1} \right)}^3}}} + 3.6 \times {10^{ - 5}}$

$= 2.56 \times {10^{ - 4}}\,\,wb/{m^2}$
4

### AIEEE 2012

A charge $Q$ is uniformly distributed over the surface of non-conducting disc of radius $R.$ The disc rotates about an axis perpendicular to its plane and passing through its center with an angular velocity $\omega .$ As a result of this rotation a magnetic field of induction $B$ is obtained at the center of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and very the radius of the disc then the variation of the magnetic induction at the center of the disc will be represented by the figure :
A
B
C
D

## Explanation

The magnetic field due a disc is given as

$B = {{{h_0}\omega Q} \over {2\pi R}}$ i.e., $B \propto {1 \over R}$