Question

In designing a velocity selector that uses uniform perpendicular electric and magnetic fields, you want to select positive ions of charge that are traveling perpendicular to the fields at 8.75 km s. The magnetic field available to you has a magnitude of 0.550 T. (a) What magnitude of electric field do you need? (b) Show how the two fields should be oriented relative to each other and to the velocity of the ions.

Answer 1

Answer:

Part a)

E = 4812.5 N/C

Part b)

Two field must be perpendicular to each other as well as perpendicular to the velocity of charge so that net force on the moving charge will be zero

Explanation:

For uniform electric and magnetic field we will have a charge particle moving through it such that net force on it is zero

so here we have

$qE = qvB$

magnetic force on moving charge will balanced by electrostatic force on moving charge

$v = \frac{E}{B}$

now we have

$v = 8.75 km/s$

B = 0.550 T

now we have

$E = (8.75 \times 10^3)(0.550)$

$E = 4812.5 N/C$

Part b)

Two field must be perpendicular to each other as well as perpendicular to the velocity of charge so that net force on the moving charge will be zero