 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2008

A horizontal overhead powerline is at height of $4m$ from the ground and carries a current of $100A$ from east to west. The magnetic field directly below it on the ground is
$\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,\,Tm\,\,{A^{ - 1}}} \right)$
A
$2.5 \times {10^{ - 7}}\,T$ southward
B
$5 \times {10^{ - 6}}\,T$ northward
C
$5 \times {10^{ - 6}}\,T$ southward
D
$2.5 \times {10^{ - 7}}\,T$ northward

Explanation

The magnetic field is

$B = {{{\mu _0}} \over {4\pi }}{{2I} \over r}$

$= {10^{ - 7}} \times {{2 \times 100} \over 4}$

$= 5 \times {10^{ - 6}}T$ According to right hand palm rule, the magnetic field is directed towards south.
2

AIEEE 2008

Relative permittivity and permeability of a material ${\varepsilon _r}$ and ${\mu _r},$ respectively. Which of the following values of these quantities are allowed for a diamagnetic material?
A
${\varepsilon _r} = 0.5,\,\,{\mu _r} = 1.5$
B
${\varepsilon _r} = 1.5,\,\,{\mu _r} = 0.5$
C
${\varepsilon _r} = 0.5,\,\,{\mu _r} = 0.5$
D
${\varepsilon _r} = 1.5,\,\,{\mu _r} = 1.5$

Explanation

For a diamagnetic material, the value of ${\mu _r}$ is less than one. For any material, the value of ${ \in _r}$ is always greater than $1.$
3

AIEEE 2007

Two identical conducting wires $AOB$ and $COD$ are placed at right angles to each other. The wire $AOB$ carries an electric current ${I_1}$ and $COD$ carries a current ${I_2}$. The magnetic field on a point lying at a distance $d$ from $O$, in a direction perpendicular to the plane of the wires $AOB$ and $COD$ , will be given by
A
${{{\mu _0}} \over {2\pi d}}\left( {I_1^2 + I_2^2} \right)$
B
${{{\mu _0}} \over {2\pi }}{\left( {{{{I_1} + {I_2}} \over d}} \right)^{{1 \over 2}}}$
C
${{{\mu _0}} \over {2\pi d}}{\left( {I_1^2 + I_2^2} \right)^{{1 \over 2}}}$
D
${{{\mu _0}} \over {2\pi d}}\left( {{I_1} + {I_2}} \right)$

Explanation

Clearly, the magnetic fields at a point $P,$ equidistant from $AOB$ and $COD$ will have directions perpendicular to each other, as they are placed normal to each other.

$\therefore$ Resultant field, $B = \sqrt {B_1^2 + B_2^2}$

But ${B_1} = {{{\mu _0}{I_1}} \over {2\pi d}}$ and ${B_2} = {{{\mu _0}{I_2}} \over {2\pi d}}$

$\therefore$ $B = \sqrt {{{\left( {{{{\mu _0}} \over {2\pi d}}} \right)}^2}\left( {I_1^2 + I_2^2} \right)}$

or, $B = {{{\mu _0}} \over {2\pi d}}{\left( {I_1^2 + I_2^2} \right)^{1/2}}$ 4

AIEEE 2007

A long straight wire of radius $a$ carries a steady current $i.$ The current is uniformly distributed across its cross section. The ratio of the magnetic field at $a/2$ and $2a$ is
A
$1/2$
B
$1/4$
C
$4$
D
$1$

Explanation

Here, current is uniformly distributed across the cross-section of the wire, therefore, current enclosed in the ampere-an path formed at a distance ${r_1}\left( { = {a \over 2}} \right)$

$= \left( {{{\pi r_1^2} \over {\pi {a^2}}}} \right) \times I,$ where $I$ is total current

$\therefore$ Magnetic field at ${P_1}$ is

${B_1} = {{{\mu _0} \times current\,\,enclosed} \over {path}}$

$\Rightarrow {B_1} = {{{\mu _0} \times \left( {{{\pi r_1^2} \over {\pi {a^2}}}} \right) \times I} \over {2\pi {r_1}}}$

$= {{{\mu _0} \times I{r_1}} \over {2\pi {a^2}}}$

Now, magnetic field at point ${P_2},$ ${B_2} = {{{\mu _0}} \over {2\pi }}.{I \over {\left( {2a} \right)}} = {{{\mu _0}I} \over {4\pi a}}$

$\therefore$ Required ratio $= {{{B_1}} \over {{B_2}}} = {{{\mu _0}I{r_1}} \over {2\pi {a^2}}} \times {{4\pi a} \over {{\mu _0}I}}$

$= {{2{r_1}} \over a} = {{2 \times {a \over 2}} \over a} = 1.$