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Chapter 19: Problem 21

An electron moves with speed \(2.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a uniform magnetic field of \(1.4 \mathrm{T},\) pointing south. At one instant, the electron experiences an upward magnetic force of $1.6 \times 10^{-14} \mathrm{N} .$ In what direction is the electron moving at that instant? 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 possibilitics.)

Short Answer

Expert verified
Answer: The electron is moving towards the West.

Step by step solution

01

Find the charge of an electron

Recall that the charge of an electron is: \(q = -1.6 \times 10^{-19} \,\text{C}\)
02

Write the formula for magnetic force

Write the formula for the magnitude of the magnetic force acting on a charged particle: \(|\vec{F}| = q|\vec{v}| |\vec{B}| \sin\theta\)
03

Plug in the given values

Replace the magnitudes of the force, charge, velocity, and the magnetic field with the given values: \(1.6 \times 10^{-14} \,\text{N} = (-1.6 \times 10^{-19} \,\text{C} ) (2.0 \times 10^{5} \frac{\text{m}}{\text{s}}) (1.4 \, \text{T}) \sin\theta\)
04

Solve for theta

Rearrange and solve for \(\theta\): \(\sin\theta = \frac{1.6 \times 10^{-14} \,\text{N}}{ (-1.6 \times 10^{-19} \,\text{C} ) (2.0 \times 10^{5} \frac{\text{m}}{\text{s}}) (1.4 \, \text{T})}\) Calculate the value of \(\sin\theta\): \(\sin\theta = \frac{1.6 \times 10^{-14}}{(-1.6 \times 10^{-19})(2.0 \times 10^5)(1.4)} = -1\) Now find the angle \(\theta\): \(\theta = \arcsin (-1)\)
05

Determine all possible angles

For a sine function, \(-1\) occurs at \(270^{\circ}\) or \(7 \pi / 2\). Therefore, \(\theta = 270^{\circ}\) or \(7 \pi / 2\). Since \(\theta\) is \(270^{\circ}\), it means the direction of the electron's velocity is \(270^{\circ}\) from south, which is towards west. So, the electron is moving towards the West at that instant.

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Most popular questions from this chapter

A straight wire is aligned east-west in a region where Earth's magnetic field has magnitude \(0.48 \mathrm{mT}\) and direction \(72^{\circ}\) below the horizontal, with the horizontal component directed due north. The wire carries a current \(I\) toward the west. The magnetic force on the wire per unit length of wire has magnitude \(0.020 \mathrm{N} / \mathrm{m}\) (a) What is the direction of the magnetic force on the wire? (b) What is the current \(I ?\)
A straight wire is aligned north-south in a region where Earth's magnetic field \(\overrightarrow{\mathbf{B}}\) is directed \(58.0^{\circ}\) above the horizontal, with the horizontal component directed due north. The wire carries a current of \(8.00 \mathrm{A}\) toward the south. The magnetic force on the wire per unit length of wire has magnitude $2.80 \times 10^{-3} \mathrm{N} / \mathrm{m} .$ (a) What is the direction of the magnetic force on the wire? (b) What is the magnitude of \(\mathbf{B} ?\)
Electrons in a television's CRT are accelerated from rest by an electric field through a potential difference of \(2.5 \mathrm{kV} .\) In contrast to an oscilloscope, where the electron beam is deflected by an electric ficld, the beam is deflected by a magnetic field. (a) What is the specd of the electrons? (b) The beam is deflected by a perpendicular magnetic field of magnitude $0.80 \mathrm{T}$. What is the magnitude of the acceleration of the electrons while in the field? (c) What is the speed of the electrons after they travel 4.0 mm through the magnetic field? (d) What strength electric field would give the electrons the same magnitude acceleration as in (b)? (c) Why do we have to use an clectric ficld in the first place to get the electrons up to speed? Why not use the large acceleration due to a magnetic field for that purpose?
The magnetic field in a cyclotron is \(0.50 \mathrm{T}\). What must be the minimum radius of the dees if the maximum proton speed desired is $1.0 \times 10^{7} \mathrm{m} / \mathrm{s} ?$
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ In one type of mass spectrometer, ions having the same velocity move through a uniform magnetic field. The spectrometer is being used to distinguish $^{12} \mathrm{C}^{+}\( and \)^{14} \mathrm{C}^{+}$ ions that have the same charge. The \(^{12} \mathrm{C}^{+}\) ions move in a circle of diameter \(25 \mathrm{cm} .\) (a) What is the diameter of the orbit of \(^{14} \mathrm{C}^{+}\) ions? (b) What is the ratio of the frequencies of revolution for the two types of ion?
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