Ncert Exercises & Solutions

4.19. 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 30°C with the initial velocity.

Magnetic field strength, B=0.15 T

Charge on the electron,

e = 1.6 \times 10^{- 19} C

Mass of the electron,

m = 9.1 \times 10^{- 31} kg

Potential difference,

V = 2.0 kV = 2 \times 10^{3} V

Thus, kinetic energy of the electron=eV

\Rightarrow eV = \frac{1}{2} {mv}^{2}

v = \sqrt{\frac{2 eV}{m}} . . . . . . . . . . . (i)

Where, v=velocity of the electron

(a) Magnetic force on the electron provides the required centripetal force of the electron.

Hence, the electron traces a circular path or radius r.

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

Centripetal force \frac{{mv}^{2}}{r}

∴ Bev = \frac{{mv}^{2}}{r}

r = \frac{mv}{Be} . . . . . . . . . . (ii)

From equation (i) and (ii), we get

r = \frac{m}{Be} {[\frac{2 eV}{m}]}^{\frac{1}{2}}

= \frac{9.1 \times 10^{- 31}}{0.15 \times 1.6 \times 10^{- 19}} \times {(\frac{2 \times 1.6 \times 10^{- 19} \times 2 \times 10^{3}}{9.1 \times 10^{- 31}})}^{\frac{1}{2}}

= 100.55 \times 10^{- 5}

= 1.01 \times 10^{- 3} m

= 1 mm

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

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

v_{1} = v \sin θ

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

r_{1} = \frac{{mv}_{1}}{Be}

= \frac{mvsinθ}{Be}

= \frac{9.1 \times 10^{- 31}}{0.15 \times 1.6 \times 10^{- 19}} \times {[\frac{2 \times 1.6 \times 10^{- 19} \times 2 \times 10^{3}}{9 \times 10^{- 31}}]}^{\frac{1}{2}} \times \sin 30 °

= 0.5 \times 10^{- 3} m

= 0.5 mm

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

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