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A magnetic field set up using Helmholtz coils 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 the 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?
Solution
Magnetic field, $B=0.75 \,T$ Accelerating voltage, $V =15\, kV =15 \times 10^{3} \,V$
Electrostatic field, $E=9 \times 10^{5} \,Vm ^{-1}$
Mass of the electron $=m$ Charge of the electron $=e$ Velocity of the electron $=v$ Kinetic energy of the electron $=e V$ $\Rightarrow \frac{1}{2} m v^{2}=e V$
$\therefore \frac{e}{m}=\frac{v^{2}}{2 V}\dots(i)$
since the particle remains undeflected by electric and magnetic fields, we can infer that the electric field is balancing the magnetic field.
$\therefore e E=e v B$
$v=\frac{E}{B}\ldots(ii)$
Putting equation $(ii)$ in equation $(i),$ we get
$\frac{e}{m}=\frac{1}{2} \frac{\left(\frac{E}{B}\right)^{2}}{V}=\frac{E^{2}}{2 V B^{2}}$
$=\frac{\left(9.0 \times 10^{5}\right)^{2}}{2 \times 15000 \times(0.75)^{2}}=4.8 \times 10^{7} \,C / kg$
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 $H e^{++}, L i^{+++}$
