Problem 7: In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.095 T magnetic field.

Complete question

In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.095 T magnetic field.

a. What electric field strength, in volts per mater, is needed to select a speed of 4.2 x 10^6 m/s?

b. What is the voltage, in kilovolts, between the plates if they are separated by 0.95 cm?

Explanation:

Given that,

magnetic field B = 0.095T

Speed of particle v = 4.2 ×10^6m/s

Separation between plate d = 0.95cm

d = 0.95/100 = 0.0095m

a. Using the mass spectrometer velocity selector relationship between the electric field and magnetic field.

v = E/B

Where

v is the speed selector

B is magnetic field

E is electric field

Therefore, E = vB

E = 4.2 × 10^6 × 0.095

E = 0.399× 10^6

E = 3.99 × 10^5 V/m

b. Voltage?

The relationship between electric field and potential difference between the two plates is given as

V = Ed

V = 3.99 × 10^5 × 0.0095

V = 3790.5 V

To kV, 1kV = 1000V

Then, V = 3.7905kV

V ≈ 3.791 kV

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2