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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, \(\theta = 90掳\), and \(\sin{90掳}=1\). 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

Which of the following has the largest cyclotron frequency? a) an electron with speed \(v\) in a magnetic field with magnitude \(B\) b) an electron with speed \(2 v\) in a magnetic field with magnitude \(B\) c) an electron with speed \(v / 2\) in a magnetic field with magnitude \(B\) d) an electron with speed \(2 v\) in a magnetic field with magnitude \(B / 2\) e) an electron with speed \(v / 2\) in a magnetic field with magnitude \(2 B\)

The figure shows a schematic diagram of a simple mass spectrometer, consisting of a velocity selector and a particle detector and being used to separate singly ionized atoms \(\left(q=+e=1.60 \cdot 10^{-19} \mathrm{C}\right)\) of gold \((\mathrm{Au})\) and molybdenum (Mo). The electric field inside the velocity selector has magnitude \(E=1.789 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\) and points toward the top of the page, and the magnetic field has magnitude \(B_{1}=1.00 \mathrm{~T}\) and points out of the page. a) Draw the electric force vector, \(\vec{F}_{E},\) and the magnetic force vector, \(\vec{F}_{B},\) acting on the ions inside the velocity selector. b) Calculate the velocity, \(v_{0}\), of the ions that make it through the velocity selector (those that travel in a straight line). Does \(v_{0}\) depend on the type of ion (gold versus molybdenum), or is it the same for both types of ions? c) Write the equation for the radius of the semicircular path of an ion in the particle detector: \(R=R\left(m, v_{0}, q, B_{2}\right)\). d) The gold ions (represented by the black circles) exit the particle detector at a distance \(d_{2}=40.00 \mathrm{~cm}\) from the entrance slit, while the molybdenum ions (represented by the gray circles) exit the particle detector at a distance \(d_{1}=19.81 \mathrm{~cm}\) from the entrance slit. The mass of a gold ion is \(m_{\text {gold }}=\) \(3.27 \cdot 10^{-25}\) kg. Calculate the mass of a molybdenum ion.

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?

An electron (with charge \(-e\) and mass \(m_{\mathrm{e}}\) ) moving in the positive \(x\) -direction enters a velocity selector. The velocity selector consists of crossed electric and magnetic fields: \(\vec{E}\) is directed in the positive \(y\) -direction, and \(\vec{B}\) is directed in the positive \(z\) -direction. For a velocity \(v\) (in the positive \(x\) -direction \(),\) the net force on the electron is zero, and the electron moves straight through the velocity selector. With what velocity will a proton (with charge \(+e\) and mass \(m_{\mathrm{p}}=1836 \mathrm{~m}_{\mathrm{e}}\) ) move straight through the velocity selector? a) \(v\) b) \(-v\) c) \(v / 1836\) d) \(-v / 1836\)

A 30 -turn square coil with a mass of \(0.250 \mathrm{~kg}\) and a side length of \(0.200 \mathrm{~m}\) is hinged along a horizontal side and carries a 5.00 -A current. It is placed in a magnetic field pointing vertically downward and having a magnitude of \(0.00500 \mathrm{~T}\). Determine the angle that the plane of the coil makes with the vertical when the coil is in equilibrium. Use \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).
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