$A$ 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?

(D) Magnetic field,$B = 0.75 \;T$. Accelerating voltage,$V = 15 \;kV = 15 \times 10^{3} \;V$. Electrostatic field,$E = 9.0 \times 10^{5} \;V \,m^{-1}$.
Let the mass of the particle be $m$ and its charge be $q$. The kinetic energy gained by the particle is $qV = \frac{1}{2}mv^2$. Thus,$\frac{q}{m} = \frac{v^2}{2V} \dots (i)$.
Since the beam remains undeflected,the electric force equals the magnetic force: $qE = qvB$,which gives $v = \frac{E}{B} \dots (ii)$.
Substituting $(ii)$ into $(i)$,we get $\frac{q}{m} = \frac{E^2}{2VB^2}$.
Substituting the values: $\frac{q}{m} = \frac{(9.0 \times 10^5)^2}{2 \times 15000 \times (0.75)^2} = \frac{81 \times 10^{10}}{30000 \times 0.5625} = \frac{81 \times 10^{10}}{16875} = 4.8 \times 10^7 \;C \,kg^{-1}$.
The specific charge $\frac{q}{m} \approx 4.8 \times 10^7 \;C \,kg^{-1}$ corresponds to the deuteron $(_{1}^{2}H^+)$ or deuterium ions. The answer is not unique because the specific charge $\frac{q}{m}$ depends only on the ratio of charge to mass. Other particles like $He^{++}$ or $Li^{+++}$ have the same $\frac{q}{m}$ ratio.