### JEE Mains Previous Years Questions with Solutions

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1

### 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

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

$I=0$

$\therefore$ $B=0$
2

### AIEEE 2007

A charged particle moves through a magnetic field perpendicular to its direction. Then
A
Kinetic energy changes but the momentum is constant
B
the momentum changes but the kinetic energy is constant
C
both momentum and kinetic energy of the particle are not constant
D
both momentum and kinetic energy of the particle are constant

## Explanation

NOTE : When a charged particle enters a magnetic field at a direction perpendicular to the direction of motion, the path of the motion is circular. In circular motion the direction of velocity changes at every point (the magnitude remains constant).

Therefore, the tangential momentum will change at every point. But kinetic energy will remain constant as it is given by ${1 \over 2}\,m{v^2}$ and ${v^2}$ is the square of the magnitude of velocity which does not change.
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

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

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

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

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