Ncert Exercises & Solutions

5.22 A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me = 9.11 × 10–31 kg). [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]

The energy of an electron beam,

E = 18 k e V = 18 \times 10^{3} e V,

Charge on an electron,

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

E = 18 \times 10^{3} \times 1.6 \times 10^{- 19} J

, Magnetic field, B =0.04 G,
Mass of an electron,

m_{e} = 9.11 \times 10^{- 19} k g

Distance up to which the electron beam travels, d= 30 cm =0.3 m
We can write the kinetic energy of the electron beam as:

\begin{array}{l} E = \frac{1}{2} m v^{2} \\ v = \sqrt{\frac{2 E}{m}} \\ = \sqrt{\frac{2 \times 18 \times 10^{3} \times 1.6 \times 10^{- 19} \times 10^{- 15}}{9.11 \times 10^{- 31}}} = 0.795 \times 10^{8} m / s \end{array}

The electron beam deflects along a circular path of radius, r.
The force due to the magnetic field balances the centripetal force of the path.

\begin{array}{l} B e V = \frac{m v^{2}}{r} \\ ∴ r = \frac{m v}{B e} \\ = \frac{9.11 \times 10^{- 31} \times 0.795 \times 10^{8}}{0.4 \times 10^{- 4} \times 1.6 \times 10^{- 19}} = 11.3 m \end{array}

Let the up and down deflection of the electron beam be x=r(1-cos)

Where

θ

=Angle of declination

\begin{array}{l} \begin{aligned} \sin θ = & \frac{d}{r} \\ = \frac{0.3}{11.3} \\ θ = \sin^{- 1} \frac{0.3}{11.3} = & {1.521}^{\circ} \end{aligned} \\ And x = 11.3 (1 - \cos {1.521}^{\circ}) \\ = 0.0039 m = 3.9 mm \end{array}

Therefore, the up and down deflection of the beam is 3.9 mm.

Energy of an electron beam, E = 18 keV = 18 \times 10^{3} eV,

Charge on an electron, e = 1.6 \times 10^{- 19} C,

E = 18 \times 10^{3} \times 1.6 \times 10^{- 19} J, Magnetic field, B = 0.04 G,

Mass of an electron, m_{e} = 9.11 \times 10^{- 19} kg

Distance up to which the electron beam travels, d = 30 cm = 0.3 m
We can write the kinetic energy of the electron beam as:

\begin{array}{l} E = \frac{1}{2} {mv}^{2} \\ v = \sqrt{\frac{2 E}{m}} \\ = \sqrt{\frac{2 \times 18 \times 10^{3} \times 1 .6 \times 10^{- 19} \times 10^{- 15}}{9 .11 \times 10^{- 31}}} = 0 .795 \times 10^{8} m / s \end{array}

The electron beam deflects along a circular path of radius, r.
The force due to the magnetic field balances the centripetal force of the path.

BeV = \frac{{mv}^{2}}{r}

∴ r = \frac{mv}{Be}

= \frac{9.11 \times 10^{- 31} \times 0.795 \times 10^{8}}{0.4 \times 10^{- 4} \times 1.6 \times 10^{- 19}} = 11.3 m

Let the up and down deflection of the electron beam be x=r(1 - cos)

θ

Where,

θ

=Angle of declination

\sin θ = \frac{d}{r}

= \frac{0.3}{11.3}

θ = \sin^{- 1} \frac{0.3}{11.3} = 1.521 °

And x=11.3(1-cos1.521

°

)

=0.0039 m = 3.9mm

Therefore, the up and down deflection of the beam is 3.9 mm.

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