Question

A velocity selector is used in a mass spectrometer to produce a beam of charged particles of uniform velocity. Suppose the fields in a selector are given by \(\vec{E}=E_{x} \hat{x}\) and \(\vec{B}=(47.45 \mathrm{mT}) \hat{y} .\) The speed with which a charged particle can travel through the selector in the \(z\) -direction without being deflected is \(5.616 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\). What is the value of \(E_{x} ?\)

Short answer

Answer: The value of the electric field E_x required is approximately -2.66 × 10^4 V/m.

Step-by-step solution

Write down the given values

We are given: - The magnetic field \(\vec{B} = (47.45\,\text{mT})\hat{y}\) - The speed in the \(z\)-direction is \(v_z = 5.616\times10^5\,\text{m/s}\). - We need to find the value of \(E_x\).

Write the equations for the forces in terms of given values

We know that the electric force \(\vec{F}_E = q\vec{E}\) and the magnetic force \(\vec{F}_B = q\vec{v}\times\vec{B}\). For a charged particle to travel without being deflected, the forces should cancel each other out: \(\vec{F}_E = \vec{F}_B\). In our case, since \(\vec{E} = E_x\hat{x}\) and \(\vec{v} = v_z\hat{z}, \vec{B} = B_y\hat{y}\), we get: - Electric force: \(\vec{F}_E = qE_x\hat{x}\) - Magnetic force: \(\vec{F}_B = q(v_z\hat{z})\times (B_y\hat{y})\)

Compute the z-component of magnetic force

Calculate the z-component of the magnetic force using the cross product: \(\vec{F}_{Bz} = v_z B_y (-\hat{x})\).

Equate electric force and magnetic force in the x-direction

Since \(\vec{F}_E = \vec{F}_B\), equate the x-components of the electric and magnetic forces: \(qE_x = -qv_zB_y\)

Solve for \(E_x\)

Now that we've equated the x-components of the forces, we can solve for \(E_x\): \(E_x = -v_zB_y\) Substitute the given values for \(v_z\) and \(B_y\): \(E_x = -(5.616\times10^5\,\text{m/s})(47.45\times10^{-3}\,\text{T})\)

Calculate the final value of \(E_x\)

Multiply the values together to find the value of \(E_x\): \(E_x \approx -2.66\times10^4\,\text{V/m}\)