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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 0.105 T magnetic field.

Part (a) What electric field strength, in volts per meter, is needed to select a speed of 3.8 × 106 m/s?
Part (b) What is the voltage, in kilovolts, between the plates if they are separated by 0.75 cm?

User Slopapa
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1 Answer

8 votes

Answer:

a).
$3.99 * 10^5 \ v/m$

b). 2.9925 kV

Step-by-step explanation:

Given :

For mass spectrometer

The magnetic field = B

B = 0.105 T

a). Given speed, v =
$3.8 * 10^6 \ m/s$

We known


$(E)/(B)=v$


$E= 3.8 * 10^6 * 0.105$


$=3.99 * 10^5 \ v/m$

b). Now spectrometer, d = 0.75 cm


$d=0.75 * 10^(-2) \ m$

We known


$E=(V)/(d)$


$V = E* d$


$V = 3.99 * 10^5 * 0.75 * 10^(-2)$


$V = 2.9925 * 10^3 \ V$

= 2.9925 kV

User Hosar
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