Question

A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. What electric field strength is needed to select a speed of 4.00×10^6 m/s?

a) 5.00×10^3 N/C
b) 1.00×10^4 N/C
c) 2.00×10^4 N/C
d) 4.00×10^4 N/C

Answer 1

The electric field strength (E) required in a velocity selector with a 0.100-T magnetic field to select a speed of 4.00 × 10^6 m/s is 4.00 × 10^5 N/C. The voltage between plates separated by 1.00 cm to achieve this electric field is 4000 volts.

Step-by-step explanation:

To find the electric field strength (E) needed for a velocity selector in a mass spectrometer, we use the relationship that the magnetic force and electric force must be equal for a particle moving with velocity (v) that will pass through the selector undeflected. This situation is expressed as qE = qvB, where q is the charge on the particle, E is the electric field strength, v is the velocity, and B is the magnetic field strength. Since q cancels out from both sides, the required electric field strength can be found with E = vB.

For a velocity (v) of 4.00 × 106 m/s and a magnetic field (B) of 0.100 T, the electric field strength (E) needed is:

E = vB = (4.00 × 106 m/s)(0.100 T) = 4.00 × 105 N/C

For part (b), to calculate the voltage between the plates (V) separated by a distance (d), we use the relationship V = Ed.

With d = 1.00 cm = 0.01 m, the voltage is:

V = E × d = (4.00 × 105 N/C)(0.01 m) = 4.00 × 103 V, or 4000 volts.