Q1) two particles of charges +q and-q are

projected from the same point with a velocity v

in a region of uniform magnetic field B such

that the velocity vector makes an angle ‘theta’

with the magnetic field. Their masses are m

and 2m then they will meet again for the first

time at a point whose distance from the point

of projection is :

Q2) a particle of specific charge q/m is

projected from the origin of coordinates with

initial velocity u towards +x and v towards –y

Uniform electric and magnetic fields exist in

the region along +y of magnitude E & B

The particle will definitely return to the origin if

a) vB/2piE is an integer

b) vB/piE is an integer

c) (u^2 + v^2)^.5B/piE is an integer

d) uB/2piE is an integer

To solve the problem of two charged particles moving in a magnetic field, we need to analyze their motion under the influence of the Lorentz force. The two particles, with charges +q and -q, will experience forces in opposite directions due to their charges, and their different masses will also affect their trajectories. Let's break this down step by step.

Understanding the Forces at Play

When a charged particle moves through a magnetic field, it experiences a magnetic force given by the equation:

F = q(v × B)

Here, F is the magnetic force, q is the charge of the particle, v is its velocity vector, and B is the magnetic field vector. The direction of the force is perpendicular to both the velocity and the magnetic field, which means the particles will undergo circular motion.

Analyzing Each Particle's Motion

• Particle 1: Charge +q, mass m
• Particle 2: Charge -q, mass 2m

For Particle 1, the magnetic force will cause it to move in a circular path. The radius of this path can be determined using the centripetal force equation:

F = (mv²)/r

Setting the magnetic force equal to the centripetal force gives us:

q(v × B) = (mv²)/r

From this, we can derive the radius of the circular motion:

r₁ = (mv)/(qB)

For Particle 2, the same analysis applies, but we need to account for its mass:

r₂ = (2mv)/(qB)

Determining the Time Period of Motion

The time period of circular motion for each particle can be calculated using:

T = (2πr)/v

For Particle 1:

T₁ = (2π(mv)/(qB))/v = (2πm)/(qB)

For Particle 2:

T₂ = (2π(2mv)/(qB))/v = (4πm)/(qB)

Finding the Meeting Point

To find when the two particles will meet again, we need to determine the least common multiple (LCM) of their time periods:

LCM(T₁, T₂) = LCM((2πm)/(qB), (4πm)/(qB)) = (4πm)/(qB)

During this time, each particle will travel a certain distance. The distance traveled by each particle can be calculated as:

d₁ = v * T₁ and d₂ = v * T₂

Substituting the time periods:

d₁ = v * (2πm)/(qB) = (2πmv)/(qB)

d₂ = v * (4πm)/(qB) = (4πmv)/(qB)

Final Calculation

Since both particles will meet at the same point after the LCM of their time periods, we can conclude that they will meet again at a distance of:

d = (4πmv)/(qB)

Thus, the distance from the point of projection where the two particles will meet for the first time is given by this formula. This result highlights how the interplay of charge, mass, and velocity in a magnetic field leads to interesting dynamics in particle motion.