Question

An electron moving with a speed \(y_{0}\) along the positive \(x\) -axis at \(\mathrm{y}=0\) enters a region of uniform magnetic field \(\mathrm{B}^{-}=-\mathrm{B}_{0} \mathrm{k} \wedge\) which exists to the right of y-axis. The electron exits from the region after some time with the speed at co-ordinate y then.

Short answer

The electron, initially moving along the positive x-axis, enters a region with a uniform magnetic field, leading to a Lorentz force causing it to move in a circular path in the xy-plane. The radius of this circular path is given by \(r = \frac{m_ev}{|q|B}\), where \(m_e\) is the mass of the electron, v is its speed, q is its charge, and B is the magnetic field. The electron covers half of the circular path and exits the region with an angle of rotation θ = π radians. Using the angle θ and the radius r, we can find the coordinate y where the electron exits the region as \(y = r \sin(θ)\). Thus, the electron exits at coordinate \(y = 0\).

Step-by-step solution

Determine the Lorentz force acting on the electron

The Lorentz force F acting on an electron moving with a velocity v in a magnetic field B is given by: \[F = q(v \times B)\] where q is the charge of the electron. Since the electron is moving along the x-axis and the magnetic field is in the -k direction, the magnetic force will result in a centripetal force causing the electron to move in a circular path in the xy-plane.

Calculate the centripetal force and radius

The centripetal force F_c acting on an electron moving in a circular path of radius r is given by: \[F_c = \frac{m_ev^2}{r}\] where me is the mass of the electron and v is the speed. Since the Lorentz force F is acting as the centripetal force, we can equate the two forces and solve for the radius r: \[F_c = F \Rightarrow \frac{m_ev^2}{r} = q(v \times B)\] From this, we get the radius r: \[r = \frac{m_ev}{|q|B}\]

Find the angle of rotation θ and the coordinate y

Since the electron exits the region with the magnetic field, it means that it completes (or covers) half of the circular path. Assuming the electron enters the region at t=0, the angle θ covered by the electron in the given time is given by: \[θ = \frac{vt}{r}\] Since the electron covers half of the circular path, its angle of rotation θ is π radians. Now, we can use the angle θ and the radius r to find the coordinate y where the electron exits the region: \[y = r \sin(θ)\]

Calculate the final coordinate y

Substituting the angle θ and radius r into the equation for y, we get: \[y = \frac{m_ev}{|q|B} \sin(\pi)\] Since sin(π) = 0, we get: \[y = 0\] So, the electron exits from the region with the magnetic field at coordinate y = 0.