Question

6.
Two particles, each having a mass m are placed
at a separation din a uniform magnetic field Bas
shown in figure. They have opposite charges of
equal magnitude g. At timet=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 valuev. 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 y =
v 12? ODAt what instant will a collision occur
between the particle if v=2v ? (d) Suppose v=2v.
and the collision between the particles is
completely inelastic Describe the motion after the
collision
x
x
X
x
X
Ac
X
X
x
dx
*
x
B
*​

Answer 1

Explanation:

It is given that  

Two particles have the mass = m

Distance = d

• It is found that there will be equal charges of magnitude in both the particles, but the opposite polarity is equal to q.
• According to the information provided, there will be a projection of both the particles towards one another.
• Assume that there is a switch of Coulomb force between the charges.

(a) The projection speed’s maximum value is vm so that there is no coalition of the two particles:

If $d=r 1+r 2$, the two particles will not collide.  

$\Rightarrow \mathrm{ym}=\mathrm{qBd} 2 \mathrm{m} ; \mathrm{d}=2 \mathrm{mv} \mathrm{mq}$.

(b) The minimum and maximum separation between the particles if $v=\frac{v m}{m}$

Let the radius of the curved path occupied by the particles, the projection of speed is $\frac{v m}{2}$ be r.

So, the minimum separation between the particles $=(d-2 r)$

$\Rightarrow(d-2 r)=d 2$

Separation distance at maximum $=(d+2 r)$

$\Rightarrow d+d 2=3 d 2$

(c) The instant between the particles at which the impact happens when

$v=2 v m$:

At a distance $\frac{d}{2}$, the particles will hit along the horizontal path.

They collide after time t.

Along the horizontal direction, the particles velocity will endure the same.

Therefore,

$T=m 2 q B$

(d) After collision, the motion of the two particles when the collision is inelastic completely:

$v=2 v m$

Particles strike at point P.

In upward direction, both the particles will have motion and at point P.

They stick together as the collision is inelastic.

Between the centres, the distance = d

Along the horizontal direction, due to the magnetic force, the velocity does not get affected.

Velocities along the horizontal direction are opposite and equal at point P, v. So, each other is cancelled.

Along the upward vertical direction, the velocity will add up.

Along the vertical direction, the magnetic force,$F=q(2 v m) B$.

Hence, these two particles will be as a united mass and change with velocity vm.