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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 (胃) is 90 degrees.
02

Calculate the magnetic force using the formula F = qvBsin胃.

Now, we can use the formula for magnetic force on a moving charge: F = qvBsin胃 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 = \(0.50 \mathrm{T}\) 胃 = 90掳 (and sin90掳 = 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 鈮� \(-1.6 \times 10^{-19} \mathrm{C} \times 2.0 \times 10^{7} \mathrm{m} / \mathrm{s} \times 0.50 \mathrm{T}\) F 鈮� \(-1.6 \times 10^{-19} \mathrm{C} \times 1.0 \times 10^{7} \mathrm{m} / \mathrm{s} \times \mathrm{T}\) F 鈮� \(-1.6 \times 10^{-12} \mathrm{N}\) 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 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.)
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.)

A tangent galvanometer is an instrument, developed in the ninetecnth century, designed to measure current based on the deflection of a compass needle. A coil of wire in a vertical plane is aligned in the magnetic north-south direction. A compass is placed in a horizontal plane at the center of the coil. When no current flows, the compass necdle points directly toward the north side of the coil. When a current is sent through the coil, the compass needle rotates through an angle \(\theta\). Derive an equation for \(\theta\) in terms of the number of coil turns \(N,\) the coil radius \(r,\) the coil current \(I,\) and the horizontal component of Earth's field \(B_{\mathrm{H}} .\) [Hint: The name of the instrument is a clue to the result. \(]\) (IMAGE CANNOT COPY)
In a certain region of space, there is a uniform electric field \(\overrightarrow{\mathbf{E}}=3.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) directed due east and a uniform magnetic field \(\mathbf{B}=0.080\) T also directed due east. What is the electromagnetic force on an electron moving due south at \(5.0 \times 10^{6} \mathrm{m} / \mathrm{s} ?\)
A uniform magnetic field of \(0.50 \mathrm{T}\) is directed to the north. At some instant, a particle with charge \(+0.020 \mu \mathrm{C}\) is moving with velocity \(2.0 \mathrm{m} / \mathrm{s}\) in a direction \(30^{\circ}\) north of east. (a) What is the magnitude of the magnetic force on the charged particle? (b) What is the direction of the magnetic force?
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