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Chapter 19: Problem 23

An electron moves at speed \(8.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a plane perpendicular to a cyclotron's magnetic field. The magnitude of the magnetic force on the electron is \(1.0 \times 10^{-13} \mathrm{N}\) What is the magnitude of the magnetic field?

Short Answer

Expert verified
Answer: The magnitude of the magnetic field is approximately \(7.81 \times 10^{-5}\,T\).

Step by step solution

01

Write down the given variables

We are given these variables: - Speed of the electron: \(v = 8.0 \times 10^{5}\,\mathrm{m/s}\) - Magnetic force on the electron: \(F = 1.0 \times 10^{-13}\,\mathrm{N}\) - Charge of the electron: \(q = -1.6 \times 10^{-19}\,\mathrm{C}\) - Angle between the velocity vector and the magnetic field vector is \(θ = 90^{\circ} ⟹ sin(θ) = 1\)
02

Solve for the magnetic field strength

Using the magnetic force formula \(F = qvBsinθ\), we can solve for the magnetic field strength \(B\). Since \(sin(θ) =1\), the equation simplifies to \(F=qvB\). Rearrange the formula to isolate the magnetic field strength: \(B= \cfrac{F}{q \times v}\) Now, plug in the given values: \(B = \cfrac{1.0 \times 10^{-13}\,\mathrm{N}}{(-1.6 \times 10^{-19}\,\mathrm{C) \times (8.0 \times 10^{5}\,\mathrm{m/s})}}\)
03

Calculate the magnetic field strength

Now, calculate the strength of the magnetic field: \(B= \cfrac{1.0\,\times\,10^{-13}\,\mathrm{N}}{((-1.6\,\times\, 10^{-19}\mathrm{C) (8.0\,\times\, 10^{5}\, \mathrm{m/s})}} \approx \boxed{7.81\,\times\,10^{-5}\,T}\) So, the magnitude of the magnetic field is approximately \(7.81 \times 10^{-5}\,T\).

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Most popular questions from this chapter

A positron \((q=+e)\) moves at \(5.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) in a magnetic field of magnitude 0.47 T. The magnetic force on the positron has magnitude \(2.3 \times 10^{-12} \mathrm{N} .\) (a) What is the component of the positron's velocity perpendicular to the magnetic field? (b) What is the component of the positron's velocity parallel to the magnetic field? (c) What is the angle between the velocity and the field?
A magnet produces a \(0.30-\mathrm{T}\) field between its poles, directed to the east. A dust particle with charge \(q=-8.0 \times 10^{-18} \mathrm{C}\) is moving straight down at \(0.30 \mathrm{cm} / \mathrm{s}\) in this field. What is the magnitude and direction of the magnetic force on the dust particle?
Two identical bar magnets lie next to one another on a table. Sketch the magnetic field lines if the north poles are at opposite ends.
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?
A solenoid of length \(0.256 \mathrm{m}\) and radius \(2.0 \mathrm{cm}\) has 244 turns of wire. What is the magnitude of the magnetic field well inside the solenoid when there is a current of \(4.5 \mathrm{A}\) in the wire?
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