An electron beam passes through a magnetic field of 4 x 10^–3 weber/m² and an electric field of 2 x 104 Vm^–1, both acting simultaneously. The path of electron remaining undeviated, calculate the speed of the electrons. If the electric field is removed, what will be the radius of the electron path?

Solution

Given,an electron beam.
Magnetic field, B = 4 x 10^–3 weber/m²Electric field, E = 2 x 10⁴ V/m

The force on the moving electron due to electric field is equal and opposite to the force on moving electron due to magnetic field since the path of moving electron is undeviated,

Therefore,
$eE = evB$
i.e.,
$v = \frac{E}{B} = \frac{2 \times 10^{4}}{4 \times 10^{- 3}} = 5 \times 10^{6} m / s .$

If, the electric field is removed and, when electron is moving perpendicular to magnetic field, the radius r of circular path traced by electron is given by,

$r = \frac{mv}{eB} = \frac{(9.1 \times 10^{- 31}) \times (5 \times 10^{6})}{(1.6 \times 10^{- 19}) \times 4 \times 10^{- 3}}$
$= 7.11 \times 10^{- 3} m = 7.11 mm$

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

Some More Questions From Moving Charges And Magnetism Chapter

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