1

### 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,

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

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

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

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

### 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
3

### JEE Main 2019 (Online) 12th January Morning Slot

As shown in the figure, two infinitely long, identical wires are bent by 90o and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If OP = OQ = 4cm, and the magnitude of the magnetic field at O is 10–4 T, and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be ($\mu$0 = 4$\pi$ $\times$ 10–7 NA–2) : A
40 A, perpendicular into the page
B
40 A, perpendicular out of the page
C
20 A, perpendicular into the page
D
40 A, perpendicular out of the page

## Explanation

Magnetic field at 'O' will be done to 'PS' and 'QN' Only

i.e. B0 = BPS + BQN $\to$ Both inwards

Let current in each wire = i

$\therefore$  B0 = ${{{\mu _0}i} \over {4\pi d}} + {{{\mu _0}i} \over {4\pi d}}$

or      10$-$4 = ${{{\mu _0}i} \over {2\pi d}}$ = ${{2 \times {{10}^{ - 7}} \times i} \over {4 \times {{10}^{ - 2}}}}$

$\therefore$       i = 20 A
4

### JEE Main 2019 (Online) 12th January Morning Slot

A proton and an $\alpha$-particle (with their masses in the ratio of 1 : 4 and charges in the ratio of 1 : 2) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii rp : r$\alpha$ of the circular paths described by them will be ;
A
$1:\sqrt 3$
B
1 : 3
C
$1:\sqrt 2$
D
1 : 2

## Explanation

KE = q$\Delta$V

r = ${{\sqrt {2mq\Delta V} } \over {qB}}$

r $\propto$ $\sqrt {{m \over q}}$

${{{r_p}} \over {{r_ \propto }}}$ = ${1 \over {\sqrt 2 }}$