Question

In a certain area, the electric field near the surface of the Earth is given by \(\vec{E}=(-150 . \mathrm{N} / \mathrm{C}) \hat{z},\) and the Earth's magnetic field is given by \(\vec{B}=(50.0 \mu \mathrm{T}) \hat{r}_{\mathrm{N}}-(20.0 \mu \mathrm{T}) \hat{z},\) where \(\hat{z}\) is a unit vector pointing vertically upward and \(\hat{r}_{\mathrm{N}}\) is a horizontal unit vector pointing due north. What velocity, \(\vec{v},\) will allow an electron in this region to move in a straight line at constant speed?

Short answer

Upon finding the net force acting on an electron in the given electric and magnetic fields, the velocity of the electron that will allow it to move in a straight line at constant speed is -3.0 x 10^5 m/s in the eastward direction.

Step-by-step solution

Label the given information

The electric field is given as\(\vec{E} = (-150 . \,\mathrm{N} / \mathrm{C}) \hat{z}\). This means it is only in the vertical direction. The magnetic field is given as \(\vec{B}=(50.0 \mu\mathrm{T})\hat{r}_{\mathrm{N}} - (20.0 \mu\mathrm{T})\hat{z}\). This means it has both horizontal and vertical components. Let's label the velocity components as \(\vec{v} = v_x \hat{x} + v_y \hat{y} + v_z \hat{z}\), where \(\hat{x}\) points due east, \(\hat{r}_{\mathrm{N}}= -\hat{y}\) points due north, and \(\hat{z}\) points upward.

Calculate the electric force

The electric force is given as: \(\vec{F}_E = -e\vec{E}\). Substitute the given electric field \(E\) and electron charge \(e = -1.6\times10^{-19}\,\mathrm{C}\) to find the electric force: \(\vec{F}_E = (1.6\times10^{-19}\,\mathrm{C})(150\,\mathrm{N}/\mathrm{C})\hat{z} = 2.4\times10^{-17}\,\mathrm{N}\hat{z}\).

Calculate the magnetic force

The magnetic force is given as: \(\vec{F}_B = -e(\vec{v} \times \vec{B})\). Substitute the given magnetic field and components of velocity: \(\vec{F}_B = +1.6\times10^{-19}\mathrm{C} [(v_x \hat{x} + v_y \hat{y} + v_z \hat{z}) \times (50.0\, \mu\mathrm{T}\, (-\hat{y}) - 20.0\, \mu\mathrm{T}\, \hat{z})]\).

Compute the cross product and forces in each direction

Compute the cross product and the forces in each direction: \(\vec{F}_B = 1.6\times10^{-19}\mathrm{C} [(-50\,\mu\mathrm{T}\,v_x\,\hat{z}) - (50\,\mu\mathrm{T}v_z\,\hat{x}) - (20\,\mu\mathrm{T}\,v_x \,\hat{y})]\).

Set the forces equal and solve for velocities

For the electron to move in a straight line at constant speed, the net forces in each direction must be zero: \(\vec{F}_E+\vec{F}_B=0 \Rightarrow -2.4 \times 10^{-17}\, \mathrm{N}\hat{z} + 1.6\times10^{-19}\,\mathrm{C} \left[(-50\,\mu\mathrm{T}\,v_x\,\hat{z}) - (50\,\mu\mathrm{T}\,v_z\,\hat{x}) - (20\,\mu\mathrm{T}\,v_x\,\hat{y})\right]=0\). Setting each component equal to zero: - \(v_z\): \(2.4\times10^{-17}\,\mathrm{N}=-50\,\mu\mathrm{T}\,v_x \times 1.6\times10^{-19}\,\mathrm{C} \Rightarrow v_x= -3.0\times10^5\,\mathrm{m/s}\). - \(v_x\): \(0=-50\mu\mathrm{T}v_z\times 1.6\times10^{-19}\,\mathrm{C} \Rightarrow v_z=0\). - \(v_y\): \(0=-20\mu\mathrm{T}v_x\times 1.6\times10^{-19}\,\mathrm{C}\) (this is consistent with the result found for \(v_x\)).

Final Result

The velocity of the electron that will allow it to move in a straight line at constant speed is: \(\vec{v}=-3.0\times10^5\,\mathrm{m/s}\hat{x}+0\,\mathrm{m/s}\hat{y}+0\,\mathrm{m/s}\hat{z} = -3.0\times10^5\,\mathrm{m/s}\hat{x}\).

Key concepts

Electric Field

The electric field is a fundamental concept in electromagnetism, representing the force per unit charge exerted on a stationary point charge. It's a vector field, meaning it has both magnitude and direction. In practical terms, if you place a positive test charge in an electric field, it will experience a force either pulling it towards or pushing it away from the source of the field. In our exercise, the electric field is given as \(\vec{E} = (-150 \, \mathrm{N} / \mathrm{C}) \hat{z}\), indicating that it points vertically upward with a strength of 150 N/C. Imagine it as an invisible line of force that might move a charged particle in the z-direction unless counteracted by other forces.

Magnetic Field

Magnetic fields are produced by electric currents, which can be macroscopic currents in wires, or microscopic currents associated with electrons in atomic orbits. The magnetic field at any given point is specified by both a direction and a magnitude; thus, it is a vector field. For the situation at hand, the Earth's magnetic field is described by \(\vec{B}=(50.0 \mu \mathrm{T})\hat{r}_{\mathrm{N}}-(20.0 \mu \mathrm{T})\hat{z}\), where \(\hat{z}\) and \(\hat{r}_{\mathrm{N}}\) represent unit vectors in the vertical and northward horizontal directions, respectively. The negative sign in front of the second term indicates that the vertical component of the magnetic field is directed downward, opposite to the electric field's direction.

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving through an electric and a magnetic field. It is the combination of electric and magnetic forces on a point charge due to electromagnetic fields. The Lorentz force formula is given by \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\), where \(q\) represents the charge of the particle, \(\vec{E}\) is the electric field, \(\vec{v}\) is the velocity of the particle, and \(\vec{B}\) is the magnetic field. The cross product \(\vec{v} \times \vec{B}\) signifies that the magnetic part of the Lorentz force is perpendicular to both the velocity of the charge and the magnetic field.

Electron Motion in Fields

By understanding the nature of electric and magnetic fields as well as the Lorentz force, one can predict the motion of an electron in these fields. Electrons are negatively charged particles and will move according to the Lorentz force exerted on them. In an instance where the electric and magnetic forces are balanced, the electron will continue to move in a straight line at a constant speed as there is no net force acting upon it. In our exercise, achieving this equilibrium required finding a velocity \(\vec{v}\) that nullifies the Lorentz force. The calculations showed that the appropriate velocity is in the x-direction, allowing the electron to move in a straight line, despite the presence of the electromagnetic fields.