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Chapter 4: Q32E (page 914)

In the Bainbridge mass spectrometer (see Fig. 27.24), the magnetic-field magnitude in the velocity selector is 0.510 T, and ions having a speed of 1.82 * 106 m/s pass through undeflected. (a) What is the electric-field magnitude in the velocity selector? (b) If the separation of the plates is 5.20 mm, what is the potential difference between the plates??

Short Answer

Expert verified

a) The electric field magnitude in the velocity selector is 9.28脳105V/m

b) The potential difference between the two plates is 4.83脳103V

Step by step solution

01

The electric force and potential difference

The magnetic force is equal to the electric force

qvB = qE

And the potential difference across the electric field divided by the separation between the plates is equal to the electric field

E=Vd

02

Determine the electric field

(a)

The magnetic force equal to the electric force

qvB=qEE=vBThereforeE=(1.82脳106m/s)(0.510T)=9.28脳105V/m

Therefore, the electric field is9.28脳105V/m

03

Determine the potential difference between plates

The potential difference between the two plates is

E=VdTherefore,V=Ed=(9.28脳105V/m)(5.20脳10鈭�3m)=4.83脳103V


Therefore, the potential difference between the two plates is4.83脳103V

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