Question

An electron moves with speed \(2.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a \(1.2-\mathrm{T}\) uniform magnctic ficld. At one instant, the electron is moving due west and experiences an upward magnetic force of $3.2 \times 10^{-14} \mathrm{N} .$ What is the direction of the magnetic field? Be specific: give the angle(s) with respect to $\mathrm{N}, \mathrm{S}, \mathrm{E}, \mathrm{W},$ up, down. (If there is more than one possible answer, find all the possibilities.)

Short answer

Answer: The magnetic field can be acting in two possible directions: 1) 45 degrees counterclockwise from the west direction, pointing northwest. 2) 45 degrees clockwise from the east direction, pointing southeast.

Step-by-step solution

Understand the relationships between quantities

The magnetic force acting on a moving charged particle is given by the Lorentz force equation: \(F = q(v \times B)\), where \(F\) is the magnetic force, \(q\) is the charge of the electron, \(v\) is its velocity, and \(B\) is the magnetic field. In this problem, we know the magnetic force, electron velocity, and the fact that the force is upward. We need to find the direction of the magnetic field.

Rewrite the Lorentz force equation as a vector equation

The equation \(F = q(v \times B)\) can be written as \(\vec{F} = q(\vec{v} \times \vec{B})\). Notice that the cross product \(v \times B\) denotes that the force experienced by the electron is perpendicular to both its velocity and the magnetic field.

Use the given data to analyze the problem

The electron is moving due west with a magnetic force acting upward. The velocity vector can be represented as \(\vec{v} = 2.0 \times 10^{5} \hat{y}\), and the force vector is given as \(\vec{F} = 3.2 \times 10^{-14} \hat{z}\). We know that the force is perpendicular to both the electron's velocity and the magnetic field vector. Therefore, \(\vec{B} = B_x \hat{x} + B_y \hat{y} + B_z \hat{z}\).

Find the components of the magnetic field vector

Applying the right-hand rule, we can determine the relationship between the vectors. The only way for the magnetic force to act upward is if the magnetic field vector points either northwest or southeast. This means that the force vector is pointing up, the velocity vector is pointing west, and the magnetic field vector is pointing in an opposite diagonal direction (either north or south). Let \(\theta\) be the angle between the magnetic field vector and the westward direction. This will give two possibilities: 1) The magnetic field vector points northwest (angle \(\theta\) with respect to W and N). 2) The magnetic field vector points southeast (angle \(\theta\) with respect to E and S). For both cases, we will divide the magnitude of the force by the magnitude of the velocity times the charge of the electron, to find the magnitude of the magnetic field: \(B = \frac{3.2 \times 10^{-14}}{(2.0 \times 10^5)(-1.6 \times 10^{-19})} = 1.2 \ \mathrm{T}\)

Calculate the angle(s) of the magnetic field vector

For both possibilities, we can use basic vector analysis along with the magnetic field magnitude (\(1.2\mathrm{T}\)) to find the angle. 1) Northwest direction: Here, \(B_x = -B\cos{\theta}\) and \(B_y = B\sin{\theta}\). Using the Pythagorean theorem, \((B_x^2 + B_y^2) = B^2\). \((1.2\cos{\theta})^2+(-1.2\sin{\theta})^2 = 1.2^2\). Solving for \(\theta\), we find it to be \(45^\circ\) (counterclockwise) with respect to both W and N. 2) Southeast direction: In this case, \(B_x = B\cos{\theta}\) and \(B_y = -B\sin{\theta}\). Following the same procedure, we find \(\theta\) to be \(45^\circ\) (clockwise) with respect to both E and S. So, there are two possible directions for the magnetic field vector: 1) 45 degrees counterclockwise from the west direction, pointing northwest. 2) 45 degrees clockwise from the east direction, pointing southeast.