Ncert Exercises & Solutions

4.20 A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and electrostatic field. If the beam remains undeflected when the electrostatic field is $9.0 \times 10^{- 5} V m^{- 1}$ , make a simple guess as to what the beam contains. Why is the answer not unique?

Magnetic field, B=0.75 T

Accelerating voltage,

V = 15 kV = 15 \times 10^{3} V

Electrostatic field,

E = 9 \times 10^{5} V m^{- 1}

Mass of the electron=m, Charge of the electron=e, Velocity of the electron=v

Kinetic energy of the electron=eV

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

∴ \frac{e}{m} = \frac{v^{2}}{2 V} . . . . . . (i)

Since the particle remains undeflected by electric and magnetic fields, we can infer that the electric field is balancing the magnetic field.

∴ eE = evB

v = \frac{E}{B} . . . . . . . . . . . (ii)

Putting equation (ii) in equation (i), we get

\frac{e}{m} = \frac{1}{2} \frac{{(\frac{E}{B})}^{2}}{V} = \frac{E^{2}}{2 {VB}^{2}}

= \frac{{(9.0 \times 10^{5})}^{2}}{2 \times 15000 \times {(0.75)}^{2}} = 4.8 \times 10^{7} C / kg

The value of specific charge e/m is equal to the value of deuteron of deuterium ions. This is not a unique answer. Other possible answers are He⁺⁺, Li⁺⁺, etc.

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