Question

An electron is accelerated by a potential difference of 3000V and enters a region of a uniform magnetic field. As a result the electron bends along a path with a radius of curvature of 26.0 cm.Find the magnitude of the magnetic field in [µT]The speed is 32.5 Mm/s (from previous question answer)

Answer 1

Given:

The potential difference V=3000 V

The radius of curvature of the curved path of the electron, r=26.0 cm=0.26 m

The velocity of the electron, v=32.5 Mm/s=32.5×10⁶ m/s

To find:

The magnitude of the magnetic field.

Step-by-step explanation:

The magnetic force acting on the electron provides it the neccessary centripetal force to trace a curved path.

Thus the centripetal force acting on the electron is equal to the magnetic force acting on it.

Therefore,

`(mv^2)/(r)=\text{evB}`

Where m is the mass of the electron, e is the charge of the electron, and B is the magnitude of the magnetic field.

On rearranging the above equation,

`B=(mv)/(er)`

On substituting the known values,

`\begin{gathered} B=(9.1*10^(-31)*32.5*10^6)/(1.6*10^(-19)*0.26) \\ =7.1*10^(-4) \\ =710\text{ }\mu\text{T} \end{gathered}`

Final answer:

The magnitude of the magnetic field is 710 μT.