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 $9.28 \times 10^{5} V / m$

b) The potential difference between the two plates is $4.83 \times 10^{3} V$

Step by step solution

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

$E = \frac{V}{d}$

Determine the electric field

(a)

The magnetic force equal to the electric force

$q v B = q E E = v B T h e r e f o r e E = (1.82 \times 10^{6} m / s) (0.510 T) = 9.28 \times 10^{5} V / m$

Therefore, the electric field is $9 .28 \times 10^{5} V / m$

Determine the potential difference between plates

The potential difference between the two plates is

$\begin{aligned} E = \frac{V}{d} \\ Therefore, \\ V = Ed \\ = (9 .28 \times 10^{5} V / m) (5 .20 \times 10^{- 3} m) \\ = 4 .83 \times 10^{3} V \end{aligned}$

Therefore, the potential difference between the two plates is $4.83 \times 10^{3} V$

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Short Answer

Step by step solution

The electric force and potential difference

Determine the electric field

Determine the potential difference between plates

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