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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

In a certain region of space, there is a uniform electric field \(\overrightarrow{\mathbf{E}}=2.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) to the east and a uniform magnetic field \(\overline{\mathbf{B}}=0.0050 \mathrm{T}\) to the west. (a) What is the electromagnetic force on an electron moving north at \(1.0 \times 10^{7} \mathrm{m} / \mathrm{s} ?(\mathrm{b})\) With the electric and magnetic fields as specified, is there some velocity such that the net electromagnetic force on the electron would be zero? If so, give the magnitude and direction of that velocity. If not, explain briefly why not.
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.)
An electromagnetic flowmeter is used to measure blood flow rates during surgery. Blood containing Na" ions flows due south through an artery with a diameter of \(0.40 \mathrm{cm} .\) The artery is in a downward magnetic field of \(0.25 \mathrm{T}\) and develops a Hall voltage of \(0.35 \mathrm{mV}\) across its diameter. (a) What is the blood specd (in \(\mathrm{m} / \mathrm{s}\) )? (b) What is the flow rate (in \(\mathrm{m}^{3} / \mathrm{s}\) )? (c) The leads of a voltmeter are attached to diametrically opposed points on the artery to measure the Hall voltage. Which of the two leads is at the higher potential?
In an electric motor, a circular coil with 100 turns of radius $2.0 \mathrm{~cm}$ can rotate between the poles of a magnet. When the current through the coil is \(75 \mathrm{~mA},\) the maximum torque that the motor can deliver is \(0.0020 \mathrm{~N} \cdot \mathrm{m} .\) (a) What is the strength of the magnetic field? (b) Is the torque on the coil clockwise or counterclockwise as viewed from the front at the instant shown in the figure?
In an electric motor, a coil with 100 turns of radius \(2.0 \mathrm{cm}\) can rotate between the poles of a magnet. The magnetic field strength is $0.20 \mathrm{T}\(. When the current through the coil is \)50.0 \mathrm{mA},$ what is the maximum torque that the motor can deliver?
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