Question

g A velocity selector consists of crossed electric and magnetic fields. The electric field has a magnitude of 480 N/C and is in the negative z direction. What should the magnetic field (magnitude and direction) be to select a proton moving in the negative x direction with a velocity of 3.50 cross times 10 to the power of 5 m/s to go un-deflected

Answer 1

B = 1.37 mT

Step-by-step explanation:

Given that,

The magnitude of the electric field, E = 480 N/C

The speed of the proton,
`v=3.50 * 10^5\ m/s`

We need to find the magnitude of the magnetic field. In a velocity selector, the electric field is balanced by the magnetic field. So,

`qE=qvB`

Where

B is the magnetic field

`B=(E)/(v)\\\\B=(480)/(3.5* 10^5)\\\\B=1.37* 10^(-3)\ T\\\\or\\\\B =1.37\ mT`

So, the magnetic field is equal to 1.37 mT.