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Chapter 19: Problem 17

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

Expert verified
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

01

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 \(\vec{F}\).
02

Determine the velocity vector

Since the muon is moving directly down toward Earth's surface, we can represent its velocity vector as \(\vec{v} = -4.5 \times 10^7\,\hat{z} \mathrm{m} / \mathrm{s}\).
03

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 \(\vec{B} = (5.0 \times 10^{-5}\sin(55^{\circ}))\,\hat{x} + (5.0 \times 10^{-5}\cos(55^{\circ}))\,\hat{z} \,\mathrm{T}\).
04

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. \(\vec{v} \times \vec{B} = (0) \,\hat{x} + (0) \,\hat{y} + (-vB_z \sin(55^{\circ}))\,\hat{z}\).
05

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 \(|\vec{F}|\).
06

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.
07

Write the final answer

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

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Most popular questions from this chapter

A singly charged ion of unknown mass moves in a circle of radius $12.5 \mathrm{cm}\( in a magnetic field of \)1.2 \mathrm{T}$. The ion was accelerated through a potential difference of \(7.0 \mathrm{kV}\) before it entered the magnetic field. What is the mass of the ion?
The strip in the diagram is used as a Hall probe to measure magnetic fields. (a) What happens if the strip is not perpendicular to the field? Does the Hall probe still read the correct field strength? Explain. (b) What happens if the ficld is in the plane of the strip?
A charged particle is accelerated from rest through a potential difference \(\Delta V\). The particle then passes straight through a velocity selector (field magnitudes \(E\) and \(B\) ). Derive an expression for the charge-to-mass ratio \((q / m)\) of the particle in terms of \(\Delta V, E,\) and \(B\)
A strip of copper \(2.0 \mathrm{cm}\) wide carries a current \(I=\) $30.0 \mathrm{A}\( to the right. The strip is in a magnetic field \)B=5.0 \mathrm{T}$ into the page. (a) What is the direction of the average magnetic force on the conduction electrons? (b) The Hall voltage is \(20.0 \mu \mathrm{V}\). What is the drift velocity?
An electromagnetic flowmeter is used to measure blood flow rates during surgery. Blood containing ions (primarily Na") flows through an artery with a diameter of \(0.50 \mathrm{cm} .\) The artery is in a magnetic field of $0.35 \mathrm{T}\( and develops a Hall voltage of \)0.60 \mathrm{mV}$ across its diameter. (a) What is the blood speed (in \(\mathrm{m} / \mathrm{s}\) )? (b) What is the flow rate (in \(\mathrm{m}^{3} / \mathrm{s}\) ) \(?\) (c) If the magnetic field points west and the blood flow is north, is the top or bottom of the artery at the higher potential?
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