Question

An electron is moving with a speed of 3.2 × 10⁷ m/s in a magnetic field of 6.00 × 10^-4 T perpendicular to its path. What will be the radium of the path? What will be frequency and the energy in keV?
[Given: mass of electron = 9.1 × 10⁻³¹ kg, charge e = 1.6 × 10⁻¹⁹ C, 1 eV = 1.6 × 10⁻¹⁹ J]

Answer 1

Given: v = 3 × 10^-7 m/s, B = 6 x 10^-4 T,

m_e = 9 x 10^-31 kg, e = 1.6 x 10^-19 C,

1 eV = 1.6 x 10^-19 J

The radius of the circular path,

r = `("m"_"e""v")/(|"e"|"B")`

`= ((9 xx 10^-31)(3 xx 10^7))/((1.6 xx 10^-19)(6 xx 10^-4)) = 2.7/9.6` = 0.2812 m

The frequency of revolution,

f = `(|"e"|"B")/(2pi"m"_"e")`

`= ((1.6 xx 10^-19)(6 xx 10^-4))/(2 xx 3.142 xx (9 xx 10^-31))`

`= 9.6/(18 xx 3.142) xx 10^8 = 16.97`MHz

Since the magnetic force does not change the kinetic energy of the charge,

KE = `1/2 "m"_"e" "v"^2 = 1/2 (9 xx 10^-31)(3 xx 10^7)^2 = 81/2 xx 10^-17`J

`= 81/(2(1.6 xx 10^-19)) xx 10^-17` eV

`= 8.1/3.2 xx 10^3`

= 2.531 ke V