Question

An electron (with charge \(-e\) and mass \(m_{\mathrm{e}}\) ) moving in the positive \(x\) -direction enters a velocity selector. The velocity selector consists of crossed electric and magnetic fields: \(\vec{E}\) is directed in the positive \(y\) -direction, and \(\vec{B}\) is directed in the positive \(z\) -direction. For a velocity \(v\) (in the positive \(x\) -direction), the net force on the electron is zero, and the electron moves straight through the velocity selector. With what velocity will a proton (with charge \(+e\) and mass \(m_{\mathrm{p}}=1836 \mathrm{~m}_{\mathrm{e}}\) ) move straight through the velocity selector? a) \(v\) b) \(-v\) c) \(v / 1836\) d) \(-v / 1836\)

Short answer

Based on the analysis and solution, the velocity of a proton that would experience no net force in the same velocity selector is the same magnitude as the electron's velocity but in the opposite direction. Hence, the correct answer is: b) -v

Step-by-step solution

Determine the relationship between electric E and magnetic B fields

The electron moves in the positive x-direction in the presence of electric E field (positive y-direction) and magnetic B field (positive z-direction). When the net force acting on the electron is zero, the force due to the electric field on the electron balances with the force due to the magnetic field. The electric force is given by \(\vec{F_e} = q\vec{E}\) and the magnetic force is given by \(\vec{F_m} = q\vec{v} \times \vec{B}\). Since the net force on electron is zero:$$ -eE = -evB $$Now, we can solve this equation to determine the relationship between electric E and magnetic B fields.

Calculate the velocity of the electron

From the equation obtained in step 1, we can find the velocity v of the electron:$$ v = \frac{E}{B} $$

Find the balance condition for the proton

The proton (with charge +e and mass \(1856m_e\)) moves in the same setup (electric E and magnetic B fields) and we need to find the velocity of the proton that will lead to no net force acting on it. Using a similar process as for the electron, the balance condition for the proton would be$$ +eE = +ev_pB $$Where \(v_p\) is the velocity of the proton.

Calculate the velocity of the proton

From the balance condition equation derived in step 3, we can find the velocity \(v_p\) of the proton:$$ v_p = \frac{E}{B} $$Since we need to compare the velocity of the proton with that of the electron, we can divide the velocities \(v_p\) and \(v\) to find their relationship:$$ \frac{v_p}{v} = \frac{\frac{E}{B}}{\frac{E}{B}} = 1 $$So the velocity of the proton is the same as the electron, but in the opposite direction, as the proton is positively charged and the electron is negatively charged. Therefore, the answer is: b) \(-v\)