### JEE Mains Previous Years Questions with Solutions

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1

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

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

### AIEEE 2007

A current $I$ flows along the length of an infinitely long, straight, thin walled pipe. Then
A
the magnetic field at all points inside the pipe is the same, but not zero
B
the magnetic field is zero only on the axis of the pipe
C
the magnetic field is different at different points inside the pipe
D
the magnetic field at any point inside the pipe is zero

## Explanation

There is no current inside the pipe. Therefore

$\oint {\overline B .\overline {d\ell } } = {\mu _0}I$

$I=0$

$\therefore$ $B=0$
3

### AIEEE 2007

A charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\overrightarrow E$ and $\overrightarrow B$ with a velocity $\overrightarrow v$ perpendicular to both $\overrightarrow E$ and $\overrightarrow B,$ and comes out without any change in magnitude or direction of $\overrightarrow v$. Then
A
$\overrightarrow v = \overrightarrow B \times \overrightarrow E /{E^2}$
B
$\overrightarrow v = \overrightarrow E \times \overrightarrow B /{B^2}$
C
$\overrightarrow v = \overrightarrow B \times \overrightarrow E /{B^2}$
D
$\overrightarrow v = \overrightarrow E \times \overrightarrow B /{E^2}$

## Explanation

Here, $\overrightarrow E$ and $\overrightarrow B$ are perpendicular to each other and the velocity $\overrightarrow v$ does not change; therefore

$qE = qvB \Rightarrow v = {E \over B}$

Also, $\left| {{{\overrightarrow E \times \overrightarrow B } \over {{B^2}}}} \right| = {{E\,\,B\sin \theta } \over {{B^2}}}$

$= {{E\,\,B\sin {{90}^ \circ }} \over {{B^2}}} = {E \over B} = \left| {\overrightarrow v } \right| = v$
4

### AIEEE 2006

A long solenoid has $200$ turns per $cm$ and carries a current $i.$ The magnetic field at its center is $6.28 \times {10^{ - 2}}\,\,\,Weber/{m^2}.$ Another long solenoid has $100$ turns per $cm$ and it carries a current ${i \over 3}$. The value of the magnetic field at its center is
A
$1.05 \times {10^{ - 2}}\,\,Weber/{m^2}$
B
$1.05 \times {10^{ - 5}}\,\,Weber/{m^2}$
C
$1.05 \times {10^{ - 3}}\,\,Weber/{m^2}$
D
$1.05 \times {10^{ - 4}}\,\,Weber/{m^2}$

## Explanation

${{{B_2}} \over {{B_1}}} = {{{\mu _0}{n_2}{i_2}} \over {{\mu _0}{n_1}{i_1}}}$

$\Rightarrow {{{B_2}} \over {6.28 \times {{10}^{ - 2}}}} = {{100 \times {i \over 3}} \over {200 \times i}}$

$\Rightarrow {B_2} = {{6.28 \times {{10}^{ - 2}}} \over 6}$

$= 1.05 \times {10^{ - 2}}\,\,Wb/{m^2}$