Question

An electron moving at a constant velocity, \(\vec{v}=v_{0} \hat{x},\) enters a region in space where a magnetic field is present. The magnetic field, \(\vec{B}\), is constant and points in the \(z\) -direction. What are the magnitude and the direction of the magnetic force acting on the electron? If the width of the region where the magnetic field is present is \(d\), what is the minimum velocity the electron must have in order to escape this region?

Short answer

Answer: The minimum velocity the electron must have in order to escape the magnetic field region is \(\frac{qd}{m\pi B}\).

Step-by-step solution

Calculate the magnetic force vector on the electron

To find the magnetic force vector acting on the electron, use the formula \(\vec{F} = q(\vec{v} \times \vec{B})\). Given that the electron's velocity vector is \(\vec{v}=v_{0} \hat{x}\) and the magnetic field vector is \(\vec{B}=B \hat{z}\), we can compute the cross product: \(\vec{F} = q(v_{0} \hat{x} \times B \hat{z}) = qv_{0}B(-\hat{y})\) The magnetic force on the electron has magnitude \(F = qv_{0}B\) and points in the negative y-direction.

Analyze the electron's motion under the influence of magnetic force

Since the magnetic force is always perpendicular to the electron's velocity, the electron will follow a circular path in the x-y plane. To escape the magnetic field region, the electron must complete less than half of the circular path before it reaches the edge of the region with a width of \(d\). The radius of the circle can be determined using centripetal force: \(F_{c} = qv_{0}B = m\frac{v_{0}^{2}}{r}\), where \(F_{c}\) is the centripetal force, m is the mass of the electron, and r is the radius of the circle. Solving for the radius r: \(r = \frac{mv_{0}}{qB}\)

Determine the condition for the electron to escape the magnetic field region

As mentioned before, the electron must travel less than half of the circular path in the width d to escape the magnetic field region. When half of the circular path is equal to d, we obtain the minimum velocity condition: \(d = \frac{1}{2}(2\pi r) = \pi r\) Substitute the expression for r: \(d = \pi \frac{mv_{0}}{qB}\)

Solve for the minimum velocity to escape the magnetic field region

Rearrange the equation in Step 3 to solve for the minimum velocity, \(v_{0}\): \(v_{0} = \frac{qd}{m\pi B}\) The minimum velocity the electron must have in order to escape the magnetic field region is \(\frac{qd}{m\pi B}\).