### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2005

Two thin, long, parallel wires, separated by a distance $'d'$ carry a current of $'i'$ $A$ in the same direction. They will
A
repel each other with a force of ${\mu _0}{i^2}/\left( {2\pi d} \right)$
B
attract each other with a force of ${\mu _0}{i^2}/\left( {2\pi d} \right)$
C
repel each other with a force $_0{i^2}/\left( {2\pi {d^2}} \right)$
D
attract each other with a force of ${\mu _0}{i^2}/\left( {2\pi {d^2}} \right)$

## Explanation

${F \over \ell } = {{{\mu _0}{i_1}} \over {2\pi d}} = {{{\mu _0}{i^2}} \over {2\pi d}}$

(attractive as current is in the same direction)
2

### AIEEE 2004

The materials suitable for making electromagnets should have
A
high retentivity and low coercivity
B
low retentivity and low coercivity
C
high retentivity and high coercivity
D
low retentivity and high coercivity

## Explanation

NOTE : Electro magnet should be amenable to magnetization & demagnetization.

$\therefore$ retentivity should be low & coercivity should be low
3

### AIEEE 2004

Two long conductors, separated by a distance $d$ carry current ${I_1}$ and ${I_2}$ in the same direction. They exert a force $F$ on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to $3d$. The new value of the force between them is
A
$- {{2F} \over 3}$
B
${F \over 3}$
C
$-2F$
D
$- {F \over 3}$

## Explanation

Force between two long conductor carrying current,

$F = {{{\mu _0}} \over {4\pi }}{{2{I_1}{I_2}} \over d} \times \ell$

$F' = - {{{\mu _0}} \over {4\pi }}{{2\left( {2{I_1}} \right){I_2}} \over {3d}}\ell$

$\therefore$ ${{F'} \over F} = {{ - 2} \over 3}$
4

### 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,

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

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