Question

An electron enters a region of space containing a uniform 0.0000193-T magnetic field. Its speed is 121 m/s and it enters perpendicularly to the field. Under these conditions, the electron undergoes circular motion. Find the radius r of the electron\'s path, and the frequency f of the motion.

Answer 1

Answer

Given,

Magnetic field, B = 0.0000193 T

speed, v = 121 m/s

mass of electron, m = 9.11 x 10⁻³¹ Kg

charge of electron, q = 1.6 x 10⁻¹⁹ C

radius of the electron path, r = ?

$r = \dfrac{mv}{qB}$

$r = \dfrac{9.31\times 10^{-31}\times 121}{1.6\times 10^{-19}\times 0.0000193}$

r = 3.64 x 10⁻⁵ m

We know frequency is inverse of time period

d = v t

$2 \pi r = v \times t$

$t = \dfrac{2 \pi r}{v}$

$t = \dfrac{2 \pi \times 3.64\times 10^{-5}}{121}$

t = 1.889 x 10⁻⁶ s.

now, frequency

$f = \dfrac{1}{t}$

$f = \dfrac{1}{1.889\times 10^{-6}}$

$f = 529380\ Hz$