Answered

Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

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

 

Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.