Question

An electron moves with a speed of 8.0×106m/s along the -z-axis. It enters a region where there is a uniform magnetic field B = (5.5T)i – (3.7T)j. What is the acceleration of the electron when it first enters the region of the uniform magnetic field?

Answer 1

Acceleration,
`a=9.36* 10^(18)\ m/s^2`

Step-by-step explanation:

It is given that,

Speed of electron,
`v=8* 10^6\ m/s`

Charge on an electron,
`q=1.6* 10^(-19)\ C`

Mass of electron,
`m=9.1* 10^(-31)\ kg`

Magnetic field,
`B=5.5i-3.7j`

Magnitude,
`|B|=√(5.5^2+(-3.77)^2)=6.66\ T`

Magnetic force is given by :

`F=qvB`

Also, F = ma

`a=(qvB)/(m)`

`a=(1.6* 10^(-19)* 8* 10^6* 6.66)/(9.1* 10^(-31))`

`a=9.36* 10^(18)\ m/s^2`

So, the acceleration of the electron is
`9.36* 10^(18)\ m/s^2`. Hence, this is the required solution.