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 x 10⁵ Vm^–1, make a simple guess as to what the beam contains. Why is the answer not unique?

Solution

Given,
Magnetic field set up, B= 0.75 T
Electrostatic field, E = 9.0 $\times$ 10⁵ V/m

Since, the magnetioc field is perpendicular to the initial velocity of electron, therefore the electron will move in a circular path.

Kinetic energy of the electron

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

$\frac{e}{m} = \frac{1.2 \times 10^{6} \times 1.2 \times 10^{6}}{2 \times 15 \times 10^{3}} \frac{e}{m} = \frac{0.24}{5} \times 10^{9} \frac{e}{m} = 0.048 \times 10^{9} \frac{e}{m} = 4.8 \times 10^{7} coulomb / kg$

This charge to mass ratio is equivalent to charge to mass ratio of proton so the charge particle may be deuterons.
However, the answer is not unique.
This is because ${He}^{+ +} (\frac{2 e}{2 m})$ and ${Li}^{+ +} (\frac{3 e}{3 m})$ have also the same value of $\frac{e}{m}$ .

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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Moving Charges And Magnetism

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