Answer:
B = 1.37 mT
Explanation:
Given that,
The magnitude of the electric field, E = 480 N/C
The speed of the proton, [tex]v=3.50 \times 10^5\ m/s[/tex]
We need to find the magnitude of the magnetic field. In a velocity selector, the electric field is balanced by the magnetic field. So,
[tex]qE=qvB[/tex]
Where
B is the magnetic field
[tex]B=\dfrac{E}{v}\\\\B=\dfrac{480}{3.5\times 10^5}\\\\B=1.37\times 10^{-3}\ T\\\\or\\\\B =1.37\ mT[/tex]
So, the magnetic field is equal to 1.37 mT.
