Question

An electron is moving through a magnetic field whose magnitude is 9.21 × 10^-4 T. The electron experiences only a magnetic force and has an acceleration of magnitude 2.30 × 10^14 m/s^2. At a certain instant, it has a speed of 7.69 × 10^6 m/s. Determine the angle (less than 90°) between the electron's velocity and the magnetic field.

Answer 1

`\theta=10.60^(\circ)`

Step-by-step explanation:

Given that,

Magnetic field,
`B=9.21* 10^(-4)\ T`

Acceleration of the electron,
`a=2.3* 10^(14)\ m/s^2`

Speed of electron,
`v=7.69* 10^(6)\ m/s`

The force due to this acceleration is balanced by the magnetic force as :

`ma=qvB\ sin\theta`

`sin\theta=(ma)/(qvB)`

`sin\theta=(9.1* 10^(-31)* 2.3* 10^(14))/(1.6* 10^(-19)* 7.69* 10^(6)* 9.21* 10^(-4))`

`\theta=10.60^(\circ)`

So, the angle between the electron's velocity and the magnetic field is 10.6 degrees. Hence, this is the required solution,