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Chapter 19: Problem 8

Find the magnetic force exerted on an electron moving vertically upward at a speed of \(2.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) by a horizontal magnetic field of \(0.50 \mathrm{T}\) directed north.

Short Answer

Expert verified
Question: Calculate the magnetic force exerted on an electron moving vertically upward with a speed of \(2.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) through a \(0.50 \mathrm{T}\) magnetic field directed north. Answer: The magnetic force on the electron is approximately \(-1.6 \times 10^{-12} \mathrm{N}\).

Step by step solution

01

Identify known values.

The values that are provided in the question are: 1. Speed of electron (v) : \(2.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) (vertically upward) 2. Magnetic field (B) : \(0.50 \mathrm{T}\) (directed north) 3. Electron charge (q) : \(-1.6 \times 10^{-19} \mathrm{C}\) (since it's an electron) Since the electron is moving vertically upward and the magnetic field is directed north (horizontal), the angle between the velocity vector and the magnetic field vector (θ) is 90 degrees.
02

Calculate the magnetic force using the formula F = qvBsinθ.

Now, we can use the formula for magnetic force on a moving charge: F = qvBsinθ Here, we already have all the known values: q = \(-1.6 \times 10^{-19} \mathrm{C}\) v = \(2.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) B = \(0.50 \mathrm{T}\) θ = 90° (and sin90° = 1) Plugging in the known values into the formula, we get: F = \((-1.6 \times 10^{-19} \mathrm{C}) (2.0 \times 10^{7} \mathrm{m} / \mathrm{s}) (0.50 \mathrm{T}) (1)\)
03

Calculate the magnetic force F.

Performing the multiplication will give us the magnetic force acting on the electron: F ≈ \(-1.6 \times 10^{-19} \mathrm{C} \times 2.0 \times 10^{7} \mathrm{m} / \mathrm{s} \times 0.50 \mathrm{T}\) F ≈ \(-1.6 \times 10^{-19} \mathrm{C} \times 1.0 \times 10^{7} \mathrm{m} / \mathrm{s} \times \mathrm{T}\) F ≈ \(-1.6 \times 10^{-12} \mathrm{N}\) The magnetic force on the electron is approximately \(-1.6 \times 10^{-12} \mathrm{N}\). The negative sign indicates that the force is in the opposite direction to the reference direction.

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Most popular questions from this chapter

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.)
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?
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ After being accelerated through a potential difference of \(5.0 \mathrm{kV},\) a singly charged carbon ion \(\left(^{12} \mathrm{C}^{+}\right)\) moves in a circle of radius \(21 \mathrm{cm}\) in the magnetic ficld of a mass spectrometer. What is the magnitude of the field?
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ Natural carbon consists of two different isotopes (excluding $^{14} \mathrm{C},$ which is present in only trace amounts). The isotopes have different masses, which is due to different numbers of neutrons in the nucleus; however, the number of protons is the same, and subsequently the chemical properties are the same. The most abundant isotope has an atomic mass of \(12.00 \mathrm{u} .\) When natural carbon is placed in a mass spectrometer, two lines are formed on the photographic plate. The lines show that the more abundant isotope moved in a circle of radius \(15.0 \mathrm{cm},\) while the rarer isotope moved in a circle of radius \(15.6 \mathrm{cm} .\) What is the atomic mass of the rarer isotope? (The ions have the same charge and are accelerated through the same potential difference before entering the magnetic field.)
An electromagnet is made by inserting a soft iron core into a solenoid. The solenoid has 1800 turns, radius \(2.0 \mathrm{cm},\) and length $15 \mathrm{cm} .\( When \)2.0 \mathrm{A}$ of current flows through the solenoid, the magnetic field inside the iron core has magnitude 0.42 T. What is the relative permeability \(\overline{K_{\mathrm{B}}}\) of the iron core? (See Section 19.10 for the definition of \(\boldsymbol{K}_{\mathbf{B}} .\) )
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