Q. 45

# Two particles, each having a mass m are placed at a separation d in a uniform magnetic filed B as shown in 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 :

Given-

Mass of two particles = *m*

Distance between the particles = *d*

Also magnitude of charges of both the particles are equal but opposite polarity equal = *q.*

From the question we can confer that, both the particles are projected towards each other with equal speed *v*.

Assuming 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-

The particles will not collide with each other if

*d* = *r*_{1} + *r*_{2}

where, *r*_{1} = *r*_{2} = radius of circular orbit followed by the charged particles

The radius of the circular path described by a particle in a magnetic field r,

where,

*m* is the mass of a proton

v= velocity of the particle

B = magnetic force

q= charge on the particle = C

So,

(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 in the condition when *v _{m}*/2, be

*r*

So, minimum separation between the particles will be

Now, the maximum distance of separation between the particles will be = (*d* + 2*r*)

(c) The instant at which the collision occurs between the particles if v=2v_{m} –

The particles along the horizontal direction will collide at a distance , *d*/2

Let they collide after time *t*.

Velocity of the particles before and after collision along the horizontal direction will remain the same.

Therefore,

From (1)

(d) The motion of the two particles after collision when the collision is completely inelastic and *v* = 2*vm* –

Let the P be t he point of particles collision.

And at point P, both the particles will have motion in upwards

As the collision is inelastic these particles will stick together.

Distance between centers = *d*

Velocity of the particles before and after collision along the horizontal direction will remain the same.

At point P, velocities along the horizontal direction are equal in magnitude and opposite in direction.

So, they will cancel out each other.

So, the velocity along the vertical upward direction will add up.

Magnetic force acting along the vertical direction,

Magnetic force, we know, Lorentz force F is given by -

where,

q = charge on an electron

v = velocity of the electron

B=magnetic field

θ= angle between B and v

Now, from Newton’s second law, the acceleration along the vertical direction,

So, from 1^{st} equation of motions-

where

v= final velocity

u=initial velocity

a= acceleration due to gravity

t=time taken

since initital velocity u =0,

Velocity of the combined mass at point P is along the vertical

direction v’.

Hence, the particles will behave as a combined mass and move with same velocity *v*_{m}.

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