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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?

Sagot :

Answer:

a).[tex]$3.99 \times 10^5 \ v/m$[/tex]

b). 2.9925 kV

Explanation:

Given :

For mass spectrometer

The magnetic field = B

B = 0.105 T

a).  Given speed, v = [tex]$3.8 \times 10^6 \ m/s$[/tex]

We known

[tex]$\frac{E}{B}=v$[/tex]

∴ [tex]$E= 3.8 \times 10^6 \times 0.105$[/tex]

      [tex]$=3.99 \times 10^5 \ v/m$[/tex]

b). Now spectrometer, d = 0.75 cm

                                      [tex]$d=0.75 \times 10^{-2} \ m$[/tex]

We known

[tex]$E=\frac{V}{d}$[/tex]

[tex]$V = E\times d$[/tex]

[tex]$V = 3.99 \times 10^5 \times 0.75 \times 10^{-2}$[/tex]

[tex]$V = 2.9925 \times 10^3 \ V$[/tex]

   = 2.9925 kV

 

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