Question

A velocity selector is used in a mass spectrometer to produce a beam of charged particles with uniform velocity. Suppose the fields in a selector are given by \(\vec{E}=\left(2.207 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\right) \hat{x}\) and \(\vec{B}=B_{y} \hat{y} .\) The speed with which charged particle can travel through the selector in the \(z\) -direction without being deflected is \(4.713 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\). What is the value of \(B_{y} ?\)

Short answer

Answer: The magnetic field component \(B_y\) is \(4.676 \cdot 10^{-2} \mathrm{T}\).

Step-by-step solution

Write the equation for the net force in the x-direction

Since the particle is not deflected, the net force in the x-direction must be zero. So, we have: $$F_{net} = F_e - F_m = 0.$$ As mentioned in the analysis, we know that \(F_e = qE\) and \(F_m = qvB\). We can, therefore, rewrite the equation as: $$qE - qvB = 0.$$

Cancel out the charge factor

We can simplify the equation further by dividing both sides by q: $$E - vB = 0.$$ Now, we can rearrange this equation to find the value of \(B\): $$B = \frac{E}{v}.$$

Substitute given values and find \(B_y\)

Lastly, substitute the given values for the electric field \(E\) and the speed \(v\) into the equation: $$B_y = \frac{2.207 \cdot 10^{4} \mathrm{~V} / \mathrm{m}}{4.713 \cdot 10^{5} \mathrm{~m} / \mathrm{s}}.$$ After calculating, we find the value of \(B_y\): $$B_y = 4.676 \cdot 10^{-2}\, \mathrm{T}.$$ The magnetic field component \(B_y\) is \(4.676 \cdot 10^{-2} \mathrm{T}\).