Question

In the Bainbridge mass spectrometer (Figure 1) , the magnetic-field magnitude in the velocity selector is

0.510 T , and ions having a speed of 1.82×106m/s pass through undeflected.

A) What is the electric-field magnitude in the velocity selector?

b) If the separation of the plates is 5.20mm, what is the potential difference between the plates?

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Answer 1

Consider the given data Magnetic field, \(B=0.510 \mathrm{~T}\) Speed of ion, \(v=1.82 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

Distance between the plates, \(d=5.20 \mathrm{~mm}=0.0052 \mathrm{~m}\)

(a)

To calculate the electric field magnitude in the velocity selector, \(E\)

Consider the relation between electric field, magnetic field and velocity

$$ \begin{array}{c} v=\frac{E}{B} \\ E=v B \end{array} $$

Substitute the values

$$ \begin{aligned} E &=\left(1.82 \times 10^{6}\right)(0.510) \\ &=0.9282 \times 10^{6} \mathrm{~N} / \mathrm{C} \end{aligned} $$

The electric field magnitude in the velocity selector is \(0.9282 \times 10^{6} \mathrm{~N} / \mathrm{C}\)

(b)

To calculate the potential difference between plates, \(V\)

Consider the relation between the distance between plates, potential difference and electric field \(v=E \times d\)

Substitute the value

$$ \begin{aligned} V &=\left(0.9282 \times 10^{6} \mathrm{~N} / \mathrm{C}\right)(0.0052 \mathrm{~m}) \\ &=0.482 \times 10^{3} \mathrm{~V} \\ &=0.482 \mathrm{kV} \end{aligned} $$

The potential difference between plates is \(0.482 \mathrm{kV}\). 3001802