Velocity of the particle just before the magnetic field is switched on is

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Answer 1

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 $x = (2 v cos θ) t_{0}$

$∴ \frac{3 m v ^{2}}{q E} = (2 v cos θ) t_{0} = (2 v cos 6 0^{\circ}) t_{0} = v t_{0}$

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

$v_{y} = u_{y} + a_{y} t_{0} = 2 v sin 6 0^{\circ} + a_{y} t_{0} = 2 v \frac{3}{2} - \frac{q E}{m} \frac{3 m v}{q E} = 3 v - 3 v = 0$

At point A, the particle velocity is purely along the x-axis and is $2 v cos 6 0^{\circ} = v$

Answer 2

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{3 m v ^{2}}{q E} = (2 v cos θ) t_{0} = (2 v cos 6 0^{\circ}) t_{0} = v t_{0}

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

v_{y} = u_{y} + a_{y} t_{0} = 2 v sin 6 0^{\circ} + a_{y} t_{0} = 2 v \frac{3}{2} - \frac{q E}{m} \frac{3 m v}{q E} = 3 v - 3 v = 0

At point A, the particle velocity is purely along the x-axis and is

2 v cos 6 0^{\circ} = v