### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2s.$ The magnet is cut along its length into three equal parts and these parts are then placed on each other with their like poles together. The time period of this combination will be
A
$2\sqrt 3 \,s$
B
${2 \over 3}\,\,s$
C
$2\,s$
D
${2 \over {\sqrt 3 }}\,s$

## Explanation

$T = 2\pi \sqrt {{1 \over {M \times B}}}$ where $I = {1 \over {12}}m{\ell ^2}$

When the magnet is cut into three pieces the pole strength will remain the same and

$M.{\rm I}.\left( {I'} \right) = {1 \over {12}}\left( {{m \over 3}} \right){\left( {{\ell \over 3}} \right)^2} \times 3 = {I \over 9}$

We have, Magnetic moment $(M)$

$=$ Pole strength $\left( m \right) \times \ell$

$\therefore$ New magnetic moment,

$M' = m \times \left( {{\ell \over 3}} \right) \times 3 = m\ell = M$

$\therefore$ $T' = {T \over {\sqrt 9 }} = {2 \over 3}s.$
2

### AIEEE 2004

The magnetic field due to a current carrying circular loop of radius $3$ $cm$ at a point on the axis at a distance of $4$ $cm$ from the centre is $54\,\mu T.$ What will be its value at the center of loop?
A
$125\,\mu T$
B
$150\,\mu T$
C
$250\,\mu T$
D
$75\,\mu T$

## Explanation

The magnetic field at a point on the axis of a circular loop at a distance $x$ from center is,

$B = {{{\mu _0}i\,{a^2}} \over {2\left( {{x^2} + {a^2}} \right)3/2}}$ $\,\,\,\,\,B' = {{{\mu _0}i} \over {2a}}$

$\therefore$ $B' = {{B.{{\left( {{x^2} + {a^2}} \right)}^{3/2}}} \over {{a^3}}}$

Put $x = 4$ & $a = 3 \Rightarrow B' = {{54\left( {{5^3}} \right)} \over {3 \times 3 \times 3}} = 250\mu T$
3

### AIEEE 2004

A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is $B.$ It is then bent into a circular loop of $n$ turns. The magnetic field at the center of the coil will be
A
$2n$ $B$
B
${n^2}\,B$
C
$nB$
D
$2{n^2}\,B$

## Explanation

KEY CONCEPT : Magnetic field at the center of a circular coil of radius $R$ carrying

current is $B = {{{\mu _0}i} \over {2R}}$

Given: $n \times \left( {2\pi r'} \right) = 2\pi R$

$\Rightarrow nr' = R\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$B' = {{n.{\mu _0}i} \over {2r'}}\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

from $\left( 1 \right)$ and $\left( 2 \right),$ $B' = {{n{\mu _0}i.n} \over {2\pi R}} = {n^2}B$
4

### AIEEE 2004

A current $i$ ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube is
A
${{{\mu _0}} \over {4\pi }},{{2i} \over r}$ tesla
B
zero
C
infinite
D
${{2i} \over r}$ tesla

## Explanation

Using Ampere's law at a distance $r$ from axis, $B$ is same from symmetry.

$\int {B.dl = {\mu _0}i}$

i.e., $B \times 2\pi r = {\mu _0}i$

Here $i$ is zero, for $r < R,$ whereas $R$ is the radius

$\therefore$ $B=0$