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Moving Charges and Magnetism — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsMoving Charges and Magnetism4 Marks
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 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 × 10–5 V m–1, make a simple guess as to what the beam contains. Why is the answer not unique?
✓

Answer

Magnetic field, B = 0.75 T
Accelerating voltage, V = 15 kV = 15 × 103 V
Electrostatic field, E = 9.0 × 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
$\Rightarrow\frac{1}{2}\text{mv}^{2}=\text{eV}$
$\therefore\frac{\text{e}}{\text{m}}=\frac{\text{v}^{2}}{2\text{V}}....(1)$
Since the particle remains undeflected by electric and magnetic fields, we can infer that the electric field is balancing the magnetic field.
$\therefore\text{eE}=\text{evB}$
$\text{v}=\frac{\text{E}}{\text{B}}...........(2)$
Putting equation (2) in equation (1), we get;
$\frac{\text{e}}{\text{m}}=\frac{1}{2}\frac{\Big(\frac{\text{E}}{\text{B}}\Big)^{2}}{\text{V}}=\frac{\text{E}^{2}}{2\text{VB}^{2}}$
$=\frac{(9.0\times10^{5})^{2}}{2\times15000\times(0.75)^{2}}$
$=4.8\times10^{7}\text{C/ kg}$
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 He++, Li++, etc.

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