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Chapter 19: Problem 65

A long straight wire carries a current of 3.2 \(A\) in the positive \(x\)-direction. An electron, traveling at \(6.8 脳 10 6\) m/s in the positive \(x\) -direction, is 4.6 cm from the wire. What force acts on the electron?

Short Answer

Expert verified
Question: Calculate the magnetic force acting on an electron moving parallel to a wire carrying a current of 3.2 A. The electron is moving at a speed of 6.8 x 10^6 m/s and is located at a distance of 4.6 cm from the wire. Answer: The magnetic force acting on the electron is approximately -3.02 x 10^(-17) N.

Step by step solution

01

Write down the known values and formula.

First, let's collect all the information given in the exercise. We have the current in the wire (\(I = 3.2 A\)), the electron's speed (\(v = 6.8 \times 10^6 m/s\)), and its distance from the wire (\(r = 4.6 cm = 0.046 m\)). The charge of the electron is \(q = -1.6 \times 10^{-19} C\). The formula to calculate the force on a moving charged particle in a magnetic field is \(F = qvB \sin{\theta}\).
02

Calculate the magnetic field strength due to the wire's current.

To find the magnetic field strength at the location of the electron, we can apply Ampere's law. The magnetic field strength (\(B\)) at a distance \(r\) from a straight wire carrying current \(I\) is described by the following formula: \(B = \frac{\mu_0 I}{2 \pi r}\), where \(\mu_0\) is the permeability of free space and has a value of \(4\pi \times 10^{-7} Tm/A\). Now, we can plug in the values we know: \(B = \frac{4\pi \times 10^{-7} \times 3.2}{2 \pi \times 0.046}\)
03

Simplify the expression for the magnetic field strength.

Now, we simplify the expression to find the magnetic field strength: \(B = \frac{4 \times 10^{-7} \times 3.2}{0.046} = 2.78 \times 10^{-5} T\)
04

Determine the angle between the velocity vector and the magnetic field.

Since the electron is moving parallel to the wire in the positive \(x\)-direction, the direction of its velocity vector is perpendicular to the magnetic field lines around the wire, which are concentric circles centered on the wire. So, the angle between the velocity vector and the magnetic field (\(\theta\)) is 90 degrees or \(\frac{\pi}{2}\) radians.
05

Calculate the magnetic force on the electron.

Now, we can use the formula for the magnetic force on a moving charged particle to find the force experienced by the electron: \(F = qvB \sin{\theta} = (-1.6 \times 10^{-19}) \times (6.8 \times 10^6) \times (2.78 \times 10^{-5}) \times \sin{\frac{\pi}{2}}\)
06

Simplify the expression for the magnetic force.

Finally, we simplify the expression to find the magnetic force on the electron: \(F = -1.6 \times 10^{-19} \times 6.8 \times 10^6 \times 2.78 \times 10^{-5} = -3.02 \times 10^{-17} N\) The force that acts on the electron is approximately \(-3.02 \times 10^{-17} N\). The negative sign indicates that the force is acting in the direction opposite to the electron's motion, which is expected due to the repulsion between the negatively charged electron and the straight wire's magnetic field lines.

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Most popular questions from this chapter

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.)
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