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.
Question:

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 $15 \mathrm{kV}$ enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is $9.0 \times 10^{5} \mathrm{~V} \mathrm{~m}^{-1}$, make a simple guess as to what the beam contains. Why is the answer not unique?

Solution:

Magnetic field, B = 0.75 T

Accelerating voltage, V = 15 kV = 15 × 103 V

Electrostatic field, E = 9 × 105 V m−1

Mass of the electron = m

Charge of the electron = e

Velocity of the electron = v

Kinetic energy of the electron = eV

$\Rightarrow \frac{1}{2} m v^{2}=e V$

$\therefore \frac{e}{m}=\frac{v^{2}}{2 V}$   …(i)

Since the particle remains undeflected by electric and magnetic fields, we can infer that the force on the charged particle due to electric field is balancing the force on the charged particle due to magnetic field.

$\therefore e E=e v B$

$v=\frac{E}{B}$    …(ii)

Putting equation (2) in equation (1), we get

$\frac{e}{m}=\frac{1}{2} \frac{\left(\frac{E}{B}\right)^{2}}{V}=\frac{E^{2}}{2 V B^{2}}$

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

This value of specific charge $e / m$ is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are $\mathrm{He}^{++} \mathrm{Li}^{++}$, etc.

 

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