Class 12 Physics

CBSE and UP Board

Moving Charges and Magnetism

Chapter-4 Exercise 4.19

Additional Exercise

# An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field 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 with the initial velocity.

Share

Magnetic field strength, B = 0.15 T

Charge on the electron, e = 1.6 x 10⁻¹

^{9}CMass of the electron, m = 9.1 x 10

^{-31}kgPotential difference, V = 2.0 kV = 2 x 10

^{3}VThus, kinetic energy of the electron = eV

=> eV = 1/2 x mv²

v = √ (2eV/m)———————–Eq-1

Where, v = velocity of the electron

Ans (a).Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron traces a circular path of radius r.

Magnetic force on the electron is given by the relation, Bev

Centripetal force mv²/r

Therefore

Bev = mv²/rr = mv/Be———————–Eq-2

From equations 1 and 2 ,we get

r = m/Be [2eV/m]

^{½}= (9.1 x 10

^{-31})/ (0.15 x1.6 x 10⁻¹^{9 }) x [(2 x 1.6 x 10⁻¹^{9 }x 2 x 10^{3})/9.1 x 10^{-31 }]^{½}= 100.55 x 10⁻⁵

= 1.01 x 10⁻³m

= 1 mm

Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.

Ans (b).When the field makes an angle 0 of 30° with initial velocity, the initial velocity will be,

v₁= vsin0

From equation (2), we can write the expression for new radius as:

r₁ =mv₁/Be = mvsin0/Be

= (9.1 x 10

^{-31})/ (0.15 x1.6 x 10⁻¹^{9 }) x [(2 x 1.6 x 10⁻¹^{9 }x 2 x 10^{3})/9.1 x 10^{-31 }]^{½ }x sin 30º= 0.5 x 10

^{-3 }m= 0.5mm

Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.