91Ó°ÊÓ

First, we will find the force due to the electric field acting on the electron using the following formula: \(\overrightarrow{\mathbf{F}}_E = q \overrightarrow{\mathbf{E}}\) Since the electric field is given as \(2.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) and the charge of an electron is \(q = -1.6 \times 10^{-19} \mathrm{C}\), we have: \(\overrightarrow{\mathbf{F}}_E = (-1.6 \times 10^{-19} \mathrm{C})(2.0 \times 10^{4} \mathrm{V} / \mathrm{m}) = -3.2 \times 10^{-15} \mathrm{N}\)

To calculate the force due to the magnetic field, use the cross product of the velocity and magnetic field: \(\overrightarrow{\mathbf{F}}_B = q(\overrightarrow{\mathbf{v}} \times \overrightarrow{\mathbf{B}})\) The electron is moving with a velocity of \(1.0 \times 10^7 \mathrm{m/s}\) to the north, and the magnetic field is \(0.0050 \mathrm{T}\) to the west. With the right-hand rule, we find that the force has an upward direction. Therefore, the magnitude of the force is: \(|\overrightarrow{\mathbf{F}}_B| = |q|(|\overrightarrow{\mathbf{v}}| |\overrightarrow{\mathbf{B}}|\sin\theta)\) Where the angle between the velocity and the magnetic field is \(\theta = 90^\circ\), and \(\sin 90^\circ = 1\). Thus, the magnitude of the magnetic force is: \(|\overrightarrow{\mathbf{F}}_B| = (1.6 \times 10^{-19} \mathrm{C})(1.0 \times 10^7 \mathrm{m/s})(0.0050 \mathrm{T}) = 8.0 \times 10^{-14} \mathrm{N}\)

Now, we can combine the electric and magnetic forces to find the net electromagnetic force acting on the electron: \(\overrightarrow{\mathbf{F}} = \overrightarrow{\mathbf{F}}_E + \overrightarrow{\mathbf{F}}_B = -3.2 \times 10^{-15} \mathrm{N} + 8.0 \times 10^{-14} \mathrm{N}\) (upward) So the electromagnetic force acting on the electron is \(8.0 \times 10^{-14} - 3.2 \times 10^{-15} = 7.68 \times 10^{-14} \mathrm{N}\) upward.

To find a velocity that results in zero net electromagnetic force, we need to find a situation where the electric and magnetic forces cancel each other out: \(\overrightarrow{\mathbf{F}}_E = -\overrightarrow{\mathbf{F}}_B\) This implies: \(|q|(\overrightarrow{\mathbf{E}}) = |q|(\overrightarrow{\mathbf{v}} \times \overrightarrow{\mathbf{B}})\) Dividing both sides by \(|q|\), we have: \(\overrightarrow{\mathbf{E}} = \overrightarrow{\mathbf{v}} \times \overrightarrow{\mathbf{B}}\) Since both the electric and magnetic fields are perpendicular to each other, we can find a velocity that makes the force due to the electric field equal and opposite to the force due to the magnetic field: \(|\overrightarrow{\mathbf{E}}| = |\overrightarrow{\mathbf{v}}||\overrightarrow{\mathbf{B}}|\sin\theta\) Solving for the velocity magnitude: \(|\overrightarrow{\mathbf{v}}| = \frac{|\overrightarrow{\mathbf{E}}|}{|\overrightarrow{\mathbf{B}}|\sin\theta} = \frac{2.0 \times 10^{4} \mathrm{V} / \mathrm{m}}{0.0050 \mathrm{T}} = 4.0 \times 10^{6} \mathrm{m/s}\) Hence, there exists a velocity with a magnitude of \(4.0 \times 10^{6} \mathrm{m/s}\) such that the net electromagnetic force on the electron would be zero. The direction of this velocity should be orthogonal to both the electric and magnetic fields. In this case, the velocity must be oriented towards the south.