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

Answer 1

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
`B= 0.0048 T`

Step-by-step explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

`\Delta V q = (1)/(2)mv^2`

Making v the subject

`v = \sqrt{[(2 \Delta V * q )/(m)] }`

Where m is the mass of electron

v is the velocity of electron

q charge on electron

`\Delta V` is the potential difference

Substituting values

`v = \sqrt{(2 * 5.9 *10^3 * 1.60218*10^(-19) )/(9.10939 *10^(-31]) )`f

`= 4.5556 *10^ {7} m/s`

For the electron to move in a circular path the magnetic force[
`F = B q v`] must be equal to the centripetal force[
`(mv^2)/(r)`] and this is mathematically represented as

`Bqv = (mv^2)/(r)`

making B the subject

`B = (mv)/(rq)`

r is the radius with a value = 5.4cm =
`= (5.4)/(100) = 5.4*10^(-2) m`

Substituting values

`B = (9.1039 *10^(-31) * 4.556 *10^7)/(5.4*10^-2 * 1.60218*10^(-19))`

`= 0.0048 T`