Question

An electron enters a region of space containing a uniform 0.0000109-T magnetic field. Its speed is 181 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

The radius of the electron's path and the frequency of the motion are
`9.44*10^(-5)\ m` and
`3.06*10^(5)\ Hz`.

Step-by-step explanation:

Given that,

Magnetic field = 0.0000109 T

Speed = 181 m/s

We need to calculate the radius of the electron

Using formula of radius

`r = (mv)/(qB)`

Where, m = mass

v = velocity

q = charge

B = magnetic field

Put the value into the formula

`r=(9.1*10^(-31)*181)/(1.6*10^(-19)* 0.0000109)`

`r=9.44*10^(-5)\ m`

We calculate the time

Using formula of time

`t=(d)/(v)`

Distance is the circumference.

`t=(2\pi r)/(v)`

Put the value into the formula

`t=(2*\pi*9.44*10^(-5))/(181)`

`t=3.27*10^(-6)\ s`

We need to calculate the frequency

The frequency is the inverse of the time for one revolution.

`f=(1)/(t)`

Put the value of t into th formula

`f=(1)/(3.27*10^(-6))`

`f=3.06*10^(5)\ Hz`

Hence, The radius of the electron's path and the frequency of the motion are
`9.44*10^(-5)\ m` and
`3.06*10^(5)\ Hz`.