Complete Question
An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?
Answer:
The magnetic field strength is [tex]B= 0.0048 T[/tex]
Explanation:
The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression
[tex]\Delta V q = \frac{1}{2}mv^2[/tex]
Making v the subject
[tex]v = \sqrt{[\frac{2 \Delta V * q }{m}] }[/tex]
Where m is the mass of electron
v is the velocity of electron
q charge on electron
[tex]\Delta V[/tex] is the potential difference
Substituting values
[tex]v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }[/tex]f
[tex]= 4.5556 *10^ {7} m/s[/tex]
For the electron to move in a circular path the magnetic force[[tex]F = B q v[/tex]] must be equal to the centripetal force[[tex]\frac{mv^2}{r}[/tex]] and this is mathematically represented as
[tex]Bqv = \frac{mv^2}{r}[/tex]
making B the subject
[tex]B = \frac{mv}{rq}[/tex]
r is the radius with a value = 5.4cm = [tex]= \frac{5.4}{100} = 5.4*10^{-2} m[/tex]
Substituting values
[tex]B = \frac{9.1039 *10^{-31} * 4.556 *10^7}{5.4*10^-2 * 1.60218*10^{-19}}[/tex]
[tex]= 0.0048 T[/tex]
