Question

What uniform magnetic field, applied perpendicular to a beam

ofelectrons moving at 1.30 x 106 m/s, is required to
makethe elctrons travel in a ciruclar arc of radius 0.350 m?

Answer 1

The magnetic field is
`2.11*10^(-5)\ T`

Step-by-step explanation:

Given that,

Speed
`v=1.30*10^(6)\ m/s`

Radius = 0.350 m

We need to calculate the magnetic field

Using formula of magnetic field

`B =(mv)/(qr)`

Where, m = mass of electron

v = speed of electron

q = charge of electron

r = radius

Put the value into the formula

`B=(9.1*10^(-31)*1.30*10^(6))/(1.6*10^(-19)*0.350)`

`B=2.11*10^(-5)\ T`

Hence, The magnetic field is
`2.11*10^(-5)\ T`