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Chapter 27: Problem 56

An electron is moving at \(v=6.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's magnetic field. If the field strength is \(0.500 \cdot 10^{-4} \mathrm{~T}\), what is the radius of the electron's circular path?

Short Answer

Expert verified
Answer: The radius of the electron's circular path is approximately \(6.83 \cdot 10^{-2}\) meters.

Step by step solution

01

Identify relevant formulas

We will use the following formulas: 1. Lorentz force: \(F = qvB\sin{\theta}\), where \(F\) is the force experienced by the charge, \(q\) is the charge, \(v\) is the velocity, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the velocity and the magnetic field. 2. Centripetal force: \(F_c = \frac{mv^2}{r}\), where \(F_c\) is the centripetal force, \(m\) is the mass of the electron, \(v\) is the velocity, and \(r\) is the radius of the circular path. Since the electron's motion is perpendicular to the magnetic field, \(\theta = 90°\), and \(\sin{90°}=1\). Thus, the Lorentz force is the centripetal force in this case.
02

Set up the equation

Now we can equate the Lorentz force \(F\) and the centripetal force \(F_c\): \(qvB = \frac{mv^2}{r}\) We want to solve for the radius \(r\).
03

Rearrange the equation

Rearrange the equation to isolate \(r\): \(r = \frac{mv}{qB}\)
04

Plug in known values

Plug in the values for the mass of an electron \(m=9.11 \cdot 10^{-31} \mathrm{kg}\), its charge \(q=-1.60 \cdot 10^{-19} \mathrm{C}\) (ignore the negative sign as we're looking for magnitude), its velocity \(v=6.00 \cdot 10^{7} \mathrm{m/s}\), and the magnetic field strength \(B=0.500 \cdot 10^{-4} \mathrm{T}\): \(r = \frac{(9.11 \cdot 10^{-31} \mathrm{kg})(6.00 \cdot 10^{7} \mathrm{m/s})}{(1.60 \cdot 10^{-19} \mathrm{C})(0.500 \cdot 10^{-4} \mathrm{T})}\)
05

Solve for the radius

Calculate the radius by solving the numerical expression: \(r = \frac{(9.11 \cdot 10^{-31})(6.00 \cdot 10^{7})}{(1.60 \cdot 10^{-19})(0.500 \cdot 10^{-4})} \approx 6.83 \cdot 10^{-2} \mathrm{m}\) Therefore, the radius of the electron's circular path is approximately \(6.83 \cdot 10^{-2}\) meters.

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Most popular questions from this chapter

In the Hall effect, a potential difference produced across a conductor of finite thickness in a magnetic field by a current flowing through the conductor is given by a) the product of the density of electrons, the charge of an electron, and the conductor's thickness divided by the product of the magnitudes of the current and the magnetic field. b) the reciprocal of the expression described in part (a). c) the product of the charge on an electron and the conductor's thickness divided by the product of the density of electrons and the magnitudes of the current and the magnetic field. d) the reciprocal of the expression described in (c). e) none of the above.

A high electron mobility transistor (HEMT) controls large currents by applying a small voltage to a thin sheet of electrons. The density and mobility of the electrons in the sheet are critical for the operation of the HEMT. HEMTs consisting of AlGaN/GaN/Si are being studied because they promise better performance at higher powers, temperatures, and frequencies than conventional silicon HEMTs can achieve. In one study, the Hall effect was used to measure the density of electrons in one of these new HEMTs. When a current of \(10.0 \mu\) A flows through the length of the electron sheet, which is \(1.00 \mathrm{~mm}\) long, \(0.300 \mathrm{~mm}\) wide, and \(10.0 \mathrm{nm}\) thick, a magnetic field of \(1.00 \mathrm{~T}\) perpendicular to the sheet produces a voltage of \(0.680 \mathrm{mV}\) across the width of the sheet. What is the density of electrons in the sheet?

A conducting rod of length \(L\) slides freely down an inclined plane, as shown in the figure. The plane is inclined at an angle \(\theta\) from the horizontal. A uniform magnetic field of strength \(B\) acts in the positive \(y\) -direction. Determine the magnitude and the direction of the current that would have to be passed through the rod to hold it in position on the inclined plane.

A proton with an initial velocity given by \((1.0 \hat{x}+$$2.0 \hat{y}+3.0 \hat{z})\left(10^{5} \mathrm{~m} / \mathrm{s}\right)\) enters a magnetic field given by \((0.50 \mathrm{~T}) \hat{z}\). Describe the motion of the proton

An electron with a speed of \(4.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) enters a uniform magnetic field of magnitude \(0.040 \mathrm{~T}\) at an angle of \(35^{\circ}\) to the magnetic field lines. The electron will follow a helical path. a) Determine the radius of the helical path. b) How far forward will the electron have moved after completing one circle?
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