X-coordinate of the particle as a of time after the magnetic field is switched on isdfrac -sqrt 3 mv-2-qE-dfrac -mv-qB-sin left -dfrac -qB-m-tright -dfrac -sqrt 3 mv-2-qE-dfrac -mv-qB-sin left -dfrac -qB-m-tright -dfrac -sqrt 3 mv-2-qE-dfrac -mv-qB-cos left -dfrac -qB-m-tright -dfrac -sqrt 3 mv-2-qE-dfrac -mv-qB-cos left -dfrac -qB-m-tright -

Question

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 $2 v cos 60^{\circ} = v$
Here, magnetic field is switched on.along the y-axis.
Its path will be helical as shown and will be with increasing pitch along -ve y-axis.
$r = \frac{m v}{q B}$
$ω = \frac{q B}{m}$
$x = x_{0} + r sin θ = \frac{\sqrt{3} m v^{2}}{q E} + \frac{m v}{q B} sin θ = \frac{\sqrt{3} m v^{2}}{q E} + \frac{m v}{q B} sin ω t$

x-coordinate of the particle as a function of time after the magnetic field is switched on is√3mv2qE−mvqBsin(qBmt)√3mv2qE+mvqBsin(qBmt)√3mv2qE−mvqBcos(qBmt)√3mv2qE+mvqBcos(qBmt)