1

### JEE Main 2019 (Online) 11th January Morning Slot

In an experiment, electrons are accelerated, from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. [Charge of the electron = 1.6 $\times$ 10–19 C Mass of the electron = 9.1 $\times$ 10–31 kg]
A
7.5 $\times$ 10$-$4 m
B
7.5 $\times$ 10$-$3 m
C
7.5 m
D
7.5 $\times$ 10$-$2 m

## Explanation

$r = {{\sqrt {2mk} } \over {eB}} = {{\sqrt {2me\Delta v} } \over {eB}}$

$r = {{\sqrt {{{2m} \over e}.\Delta v} } \over B} = {{\sqrt {{{2 \times 9.1 \times {{10}^{ - 31}}} \over {1.6 \times {{10}^{ - 19}}}}\left( {500} \right)} } \over {100 \times {{10}^{ - 3}}}}$

$r = {{\sqrt {{{9.1} \over {0.16}} \times {{10}^{ - 10}}} } \over {{{10}^{ - 1}}}} = {3 \over 4} \times {10^{ - 4}}$

$= 7.5 \times {10^{ - 4}}$
2

### JEE Main 2019 (Online) 11th January Morning Slot

There are two long co-axial solenoids of same length $l$. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2, respectively. The ratio of mutual inductance to the self - inductance of the inner-coil is :
A
${{{n_2}} \over {{n_1}}}.{{{r_2}^2} \over {{r_1}^2}}$
B
${{{n_2}} \over {{n_1}}}$
C
${{{n_1}} \over {{n_2}}}$
D
${{{n_2}} \over {{n_1}}}.{{{r_1}} \over {{r_2}}}$

## Explanation

$M = {\mu _0}\,{n_1}\,{n_2}\,\pi r_1^2$

$L = {\mu _0}\,n_1^2\,\pi r_1^2$

$\Rightarrow \,\,{M \over L} = {{{n_2}} \over {{n_1}}}$
3

### JEE Main 2019 (Online) 11th January Evening Slot

The region between y = 0 and y = d contains a magnetic field $\overrightarrow B = B\widehat z$. A particle of mass m and charge q enters the region with a velocity $\overrightarrow v = v\widehat i.$ If d $=$ ${{mv} \over {2qB}},$ the acceleration of the charged particle at the point of its emergence at the other side is :
A
${{qvB} \over m}\left( -{{{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)$
B
${{qvB} \over m}\left( {{1 \over 2}\widehat i - {{\sqrt 3 } \over 2}\widehat j} \right)$
C
${{qvB} \over m}\left( {{{ - \widehat j + \widehat i} \over {\sqrt 2 }}} \right)$
D
${{qvB} \over m}\left( {{{\widehat j + \widehat i} \over {\sqrt 2 }}} \right)$

## Explanation Here R = ${{mv} \over {qB}}$ = 2d

cos $\theta$ = ${{{R \over 2}} \over R}$ = ${1 \over 2}$

$\Rightarrow$ $\theta$ = 60o

Acceleration of the charged particle at the point of its emergence,

$\overrightarrow {{a_c}} = {a_{{c_x}}}\left( { - \widehat i} \right) + {a_{{c_y}}}\left( { - \widehat j} \right)$

= ${a_c}\cos 30^\circ \left( { - \widehat i} \right) + {a_c}\sin 30^\circ \left( { - \widehat j} \right)$

= ${a_c}\left( {{{\sqrt 3 } \over 2}\left( { - \widehat i} \right) + {1 \over 2}\left( { - \widehat j} \right)} \right)$

= ${{qvB} \over m}\left( { - {{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)$
4

### JEE Main 2019 (Online) 11th January Evening Slot

A particle of mass m and charge q is in an electric and magnetic field given by
$\overrightarrow E = 2\widehat i + 3\widehat j;\,\,\,\overrightarrow B = 4\widehat j + 6\widehat k.$

The charged particle is shifted from he origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is :
A
(2.5) q
B
(0.35) q
C
(0.15) q
D
5 q

## Explanation

${\overrightarrow F _{net}} = q\overrightarrow E + q\left( {\overrightarrow v \times \overrightarrow B } \right)$

$= \left( {2q\widehat i + 3q\widehat j} \right) + q\left( {\overrightarrow v \times \overrightarrow B } \right)$

$W = {\overrightarrow F _{net}}.\overrightarrow S$

$=$ 2q + 3q

$=$ 5q