Question

An electron of kinetic energy 1.67 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 37.4 cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.

Answer 1

Explanation:

Given that,

The kinetic energy of the electron, $K=1.67\ keV=1.67\times 10^3\ eV$

Radius of the orbit, r = 37.4 cm = 0.374 m

To find,

(a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.

Solve,

We know that, $1\ eV=1.6\times 10^{-19}\ J$

$K=1.67\times 10^3\times 1.6\times 10^{-19}\ J=2.67\times 10^{-16}\ J$

(a) The formula of the kinetic energy is given by :

$K=\dfrac{1}{2}mv^2$

v is the speed if electron

m is the mass of electron

$v=\sqrt{\dfrac{2K}{m}}$

$v=\sqrt{\dfrac{2\times 2.67\times 10^{-16}}{9.1\times 10^{-31}}}$    

$v=2.42\times 10^7\ m/s$

(b) Let B is the magnitude of magnetic field. On the circular path the magnetic field is given by :

$B=\dfrac{mv}{qr}$

$B=\dfrac{9.1\times 10^{-31}\times 2.42\times 10^7}{1.6\times 10^{-19}\times 0.374}$

$B=3.68\times 10^{-4}\ T$

(c) Let T is the time period of the motion. It is given by :

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

Circling frequency is given by :

$f=\dfrac{1}{T}=\dfrac{v}{2\pi r}$

$f=\dfrac{2.42\times 10^7}{2\pi \times 0.374 }$

$f=1.02\times 10^7\ Hz$

(d) The period of motion,

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

$T=\dfrac{1}{1.02\times 10^7}$

$T=9.80\times 10^{-8}\ s$