Velocity of the particle just before the magnetic field is switched on is
$v^i$
$v^i + \frac{\sqrt{3} v}{2}^j$
$v^i - \frac{\sqrt{3} v}{2}^j$
$2 v^i - \frac{\sqrt{3} v}{2}^j$

Question

Velocity of the particle just before the magnetic field is switched on isvhat ivhat i-dfrac -sqrt 3v-2-hat jvhat i-dfrac -sqrt 3v-2-hat j2vhat i-dfrac -sqrt 3v-2-hat j

Toppr Admin · Accepted Answer

The particle will travel in a parabolic trajectory OA.
Let the time to reach A is $t_{0}$ .

$a_{y} = - \frac{q E}{m}$

x-coordinate of the point A is $x = (2 v cos θ) t_{0}$

$∴ \frac{\sqrt{3} m v^{2}}{q E} = (2 v cos θ) t_{0} = (2 v cos 60^{\circ}) t_{0} = v t_{0}$

$\Rightarrow t_{0} = \frac{\sqrt{3} m v^{2}}{v q E} = \frac{\sqrt{3} m v}{q E}$

$v_{y} = u_{y} + a_{y} t_{0} = 2 v sin 60^{\circ} + a_{y} t_{0} = 2 v \frac{\sqrt{3}}{2} - \frac{q E}{m} \frac{\sqrt{3} m v}{q E} = \sqrt{3} v - \sqrt{3} v = 0$

At point A, the particle velocity is purely along the x-axis and is $2 v cos 60^{\circ} = v$

Velocity of the particle just before the magnetic field is switched on isv^iv^i+√3v2^jv^i−√3v2^j2v^i−√3v2^j

Velocity of the particle just before the magnetic field is switched on is
$v^i$
$v^i + \frac{\sqrt{3} v}{2}^j$
$v^i - \frac{\sqrt{3} v}{2}^j$
$2 v^i - \frac{\sqrt{3} v}{2}^j$