When an charged particle having charge q is accelerated by a potential difference V volts it gain kinetic energy given by

K = Vq, which imparts it with velocity

We know kinetic energy of a Particle with mass m kg and velocity v is given by

Equating both equations

We get velocity of the particle as

Here Potential difference of the particle, V = 2.0 kV

= 2000 V

Particle is electron, charge of electron

q = 1.6 × 10 ^-19 C

mass of electron

m = 9.1 × 10^-31 Kg

so velocity of the particle,

m/s

Solving we get

Velocity of particle,

v = m/s

(a) Now here the particle enters the magnetic field orthogonally so it will experience Lorentz magnetic force is given by :

Where is the force on the Charged particle having charge q, moving with a velocity in a magnetic Field

Where, and are magnitude of velocity and magnetic field respectively, θ is the angle between Velocity of particle and magnetic field and ŵ is a unit vector perpendicular to plane containing and

Here, since particle entered orthogonally now force on particle will be perpendicular to both magnetic field and its velocity at every instant, since magnetic field and velocity are in one plane and force is in a plane perpendicular to them, this force does not change the magnitude of velocity or speed it only changed it the direction of particle and force always act tangentially to the velocity of the particle causing it to move in a circular trajectory

As shown in the figure

Now the required centripetal force for circular motion is provided by the magnetic force we know centripetal force is given by

,

Where F is the centripetal force,

m is the mass of the particle,

moving in a circular region of radius r,

with velocity v

Magnitude of magnetic force on particle moving with velocity v in a magnetic field B orthogonally is

F = qvB(sinθ)

here at every instant so

F = qvB

Equating both equations

We get radius trajectory of the particle,

The particle is electron, thus, charge of electron is:

q = 1.6 × 10 ^-19 C

mass of electron

Kg

Velocity of particle,

v = m/s

Magnitude of magnetic field

B = 0.15 T

so

= 100.47 m

= 1.00 mm

So trajectory is a circle of radius 1.00 mm normal to plane of magnetic field

(b) when the particle enters the magnetic field, with its velocity at some angle with it, and not exactly orthogonal, particle follows a helical path, since there is force on the component of velocity which is perpendicular to the magnetic field, and No force is experience due to the component of velocity parallel to magnetic field, which keeps moving the particle forward so path is helical instead of circular. Velocity makes angle of 30 degree with the magnetic field, consider magnetic field along x axis

so particle moving in x-y plane initially making an angle 30^o with x axis, so using the Fleming left hand rule we find out the force is Toward positive Z axis,

as shown in figure

Explanation: It can be visualised as particle moving in circular path, at the same time moving forward, like an spiral, if there would have been no horizontal component of velocity it would have moved in a circular path.

Now particle tends to move in circular path in y-z plane due to Y component of velocity perpendicular to the field,and move forward along x axis due to x component of velocity,

Now angle made by velocity with x-axis,

θ = 30°

So x component of velocity,

m/s

So y component of velocity,

m/s

For the radius of the path we will again use the equation

where, q is charge of Particle which is electron so

C

m is mass of particle, i.e. mass of electron

Kg

Component of Velocity of particle undergoing circular motion,

v_y = m/s

Magnitude of magnetic field

B = 0.15 T

Putting values in equation

m

= 0.5mm

So radius of the helical path followed by electron is 0.5 mm and is moving along magnetic field with speed m/s.