Question

(b) With a current of 0.38 A, the average velocity of an electron in the wire is 5.5  10-6 m s-1 and the average magnetic force on one electron is 1.4  10-25 N. Calculate the flux density B of the magnetic field.

Answer 1

B = 0.159 T

Step-by-step explanation:

Given that,

Current, I = 0.38 A

The average velocity of the electron,
`v=5.5* 10^(-6)\ m/s`

The average magnetic force on the electron,
`F=1.4* 10^(-25)\ N`

We need to find the flux density B of the magnetic field. We know that the magnetic force is given by :

`F=qvB\\\\B=(F)/(qv)`

Put all the values,

`B=(1.4* 10^(-25))/(1.6* 10^(-19)* 5.5* 10^(-6))\\\\B=0.159\ T`

So, the required flux density of the magnetic field is 0.159 T.