Question

At a certain point on Earth's surface in the southern hemisphere, the magnetic field has a magnitude of \(5.0 \times 10^{-5} \mathrm{T}\) and points upward and toward the north at an angle of \(55^{\circ}\) above the horizontal. A cosmic ray muon with the same charge as an electron and a mass of $1.9 \times 10^{-28} \mathrm{kg}$ is moving directly down toward Earth's surface with a speed of \(4.5 \times 10^{7} \mathrm{m} / \mathrm{s} .\) What is the magnitude and direction of the force on the muon?

Short answer

Answer: After solving for the magnitude of the force using the Lorentz force formula, we calculate that the magnitude of the force is approximately \(5.76 \times 10^{-17} \, \mathrm{N}\). The force acts in the upward direction.

Step-by-step solution

Identify the knowns and unknowns

In this problem, we know the magnetic field \(\vec{B}\), the angle \(55^{\circ}\), the mass of the cosmic ray muon \(m = 1.9 \times 10^{-28} \mathrm{kg}\), and its speed \(v = 4.5 \times 10^{7} \mathrm{m} / \mathrm{s}\). The charge of the muon is the same as that of an electron, so \(q = -1.6 \times 10^{-19} \mathrm{C}\). Our goal is to find the magnitude and direction of the force \(\vec{F}\).

Determine the velocity vector

Since the muon is moving directly down toward Earth's surface, we can represent its velocity vector as \(\vec{v} = -4.5 \times 10^7\,\hat{z} \mathrm{m} / \mathrm{s}\).

Express the magnetic field vector in Cartesian coordinates

We are given that the magnetic field points upward and toward the north at an angle of \(55^{\circ}\) above the horizontal. We can break it up into horizontal and vertical components: 1. Horizontal direction: \(B_x = B\sin(55^{\circ})\) 2. Vertical direction: \(B_z = B\cos(55^{\circ})\) Thus, the magnetic field vector can be written as \(\vec{B} = (5.0 \times 10^{-5}\sin(55^{\circ}))\,\hat{x} + (5.0 \times 10^{-5}\cos(55^{\circ}))\,\hat{z} \,\mathrm{T}\).

Compute the cross product of the velocity and magnetic field vectors

Next, we need to compute the cross product \(\vec{v} \times \vec{B} = (v_xB_x - v_xB_z, 0, v_xB_x + v_xB_z)\): 1. \(\vec{v} \times \vec{B} = (0) \,\hat{x} + (0) \,\hat{y} + (-vB_z \sin(55^{\circ}))\,\hat{z}\).

Calculate the force vector and its magnitude

Now, we can find the force vector using the Lorentz force formula: \(\vec{F} = q(\vec{v} \times \vec{B}) = -1.6 \times 10^{-19} \mathrm{C} \times (-vB_z \sin(55^{\circ}))\,\hat{z} \) \(|\vec{F}| = |q|v|B_z|\sin(55^{\circ}) = (1.6 \times 10^{-19} \mathrm{C})(4.5 \times 10^7 \mathrm{m} / \mathrm{s})(5.0 \times 10^{-5}\cos(55^{\circ}))\mathrm{T}\sin(55^{\circ})\) Calculate the numerical value of \(|\vec{F}|\).

Determine the direction of the force

Since the force is nonzero only in the z-direction, and it has a positive magnitude, it acts in the upward direction.

Write the final answer

Compute the magnitude of the force \(|\vec{F}|\) and express the force as a vector in the upward direction.