Question

An electron emitted by a heated cathode and accelerated through apotential difference of 2.0 kV, enters a region with uniform magneticfield of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30º withthe initial velocity.

Answer 1

Given: The potential difference is 2 kV and the uniform magnetic field is 0.15 T.

a)

As the field is transverse to its initial velocity that is field is perpendicular to its initial velocity therefore, the electron will move in a circular path.

The required centripetal force is given by the magnetic field on the electron.

m v 2 r =Bev r= mv Be (1)

Where, r is the radius of circular path, m is the mass of electron, v is the initial velocity of electron, B is the magnetic field strength and e is the electronic charge.

The kinetic energy of the electron is given as,

1 2 m v 2 =eV v= 2eV m (2)

From equation (1) and (2), we get

r= m Be × 2eV m = 1 B 2mV e

By substituting the given values in the above equation, we get

r= 1 0.15 2×9.1× 10 −31 ×2× 10 3 1.6× 10 −19 =1× 10 −3  m =1 mm

Thus, the electron moves in a circular path of radius 1 mm.

b)

When the field makes an angle of θ with the initial velocity then electron will move in a helical path.

The velocity component in the perpendicular direction of magnetic field is,

v ⊥ =vsinθ

Then from equation (2) we get,

v ⊥ = 2eV m sinθ

Now, from equation (1), we get

r ′ = m v ⊥ Be = m Be 2eV m sinθ = 1 B 2mV e sinθ =rsinθ

By substituting the values in the above equation, we get

r ′ =( 1 mm )×sin30° =0.5 mm

Thus, the electron moves in a helical path of radius 0.5 mm.