Question

Two particles, each with mass m are placed at a separation d in a uniform magnetic field B, as shown in the figure. They have opposite charges of equal magnitude q. At time t = 0, the particles are projected towards each other, each with a speed v. Suppose the Coulomb force between the charges is switched off. (a) Find the maximum value v_m of the projection speed, so that the two particles do not collide. (b) What would be the minimum and maximum separation between the particles if v = v_m/2? (c) At what instant will a collision occur between the particles if v = 2v_m? (d) Suppose v = 2v_m and the collision between the particles is completely inelastic. Describe the motion after the collision.

Answer 1

Given,
Mass of two particles = m
Distance between them = d
Both the particles have equal charges in magnitude but opposite polarity equal to q.
As per the question, both the particles are projected towards each other with equal speed v.
It is assumed that Coulomb force between the charges is switched off.

(a) The maximum value v_m of the projection speed so that the two particles do not collide:
Both the particles will not collide if
d = r₁ + r₂ (where, r₁ = r₂ = radius of circular orbit described by the charged particles)
$$\begin{gathered} \Rightarrow d=\frac{m{v}_{\mathrm{m}}}{qB}+\frac{m{v}_{\mathrm{m}}}{qB}=\frac{2m{\mathrm{v}}_{\mathrm{m}}}{qB} \\ \mathrm{So},v_{\mathrm{m}}=\frac{qBd}{2m} \end{gathered}$$



(b) The minimum and maximum separation between the particles if v = v_m/2:
Let the radius of the curved path taken by the particles, when they are projected with speed v_m/2, be r.
So, minimum separation between the particles = (d $-$ 2r)

$$\begin{gathered} \Rightarrow (d-2r)=\frac{2m{v}_{\mathrm{m}}}{qB}-\frac{2mv}{qB} \\ \Rightarrow (d-2r)=\frac{m{v}_{\mathrm{m}}}{qB} \\ \Rightarrow (d-2r)=\frac{d}{2} \end{gathered}$$

Maximum distance separation = (d + 2r)
$\Rightarrow d+2r=d+\frac{d}{2}=\frac{3d}{2}$

(c) The instant at which the collision occurs between the particles when $v=2v_{\mathrm{m}}$:
The particles will collide at a distance d/2 along the horizontal direction.



Let they collide after time t.
Velocity of the particles along the horizontal direction will remain the same.
Therefore, $t=\frac{d/2}{2v_m}$
$$\begin{gathered} \Rightarrow t=\frac{d}{4}\times \frac{2m}{qBd} \\ \Rightarrow t=\frac{m}{2qB} \end{gathered}$$

(d) The motion of the two particles after collision when the collision is completely inelastic:
v = 2v_m
Let the particles collide at point P.
And at point P, both the particles will have motion in upward direction.
As the collision is inelastic they stick together.
Distance between centres = d
Velocity along the horizontal direction does not get affected due to the magnetic force.

At point P, velocities along the horizontal direction are equal and opposite. So, they cancel each other.

Velocity along the vertical direction (upward) will add up.
Magnetic force acting along the vertical direction,
$F=q\left(2v_m\right)B$
Acceleration along the vertical direction,
$a=\frac{F}{m}=\frac{2q{v}_{m}B}{m}$

Velocity of the combined mass at point P is along the vertical direction. So,
$$\begin{gathered} v\text{'}=0+a\times t \\ v\text{'}=0+\left(\frac{2q{v}_{m}B}{m}\right)\times \left(\frac{m}{2qB}\right) \\ v\text{'}=v_m \end{gathered}$$
Hence, both the particles will behave as a combined mass and move with velocity v_m.