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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

A circular coil with a radius of \(10.0 \mathrm{~cm}\) has 100 turns of wire and carries a current, \(i=100 . \mathrm{mA} .\) It is free to rotate in a region with a constant horizontal magnetic field given by \(\vec{B}=(0.0100 \mathrm{~T}) \hat{x}\). If the unit normal vector to the plane of the coil makes an angle of \(30.0^{\circ}\) with the horizontal, what is the magnitude of the net magnetic torque acting on the coil? 27.61 At \(t=0\) an electron crosses the positive \(y\) -axis (so \(x=0\) ) at \(60.0 \mathrm{~cm}\) from the origin with velocity \(2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. It is in a uniform magnetic field. a) Find the magnitude and the direction of the magnetic field that will cause the electron to cross the \(x\) -axis at \(x=60.0 \mathrm{~cm}\). b) What work is done on the electron during this motion? c) How long will the trip take from \(y\) -axis to \(x\) -axis?

A square loop of wire of side length \(\ell\) lies in the \(x y\) -plane, with its center at the origin and its sides parallel to the \(x\) - and \(y\) -axes. It carries a current, \(i\), in the counterclockwise direction, as viewed looking down the \(z\) -axis from the positive direction. The loop is in a magnetic field given by \(\vec{B}=\left(B_{0} / a\right)(z \hat{x}+x \hat{z}),\) where \(B_{0}\) is a constant field strength, \(a\) is a constant with the dimension of length, and \(\hat{x}\) and \(\hat{z}\) are unit vectors in the positive \(x\) -direction and positive \(z\) -direction. Calculate the net force on the loop.

An electron is moving with a constant velocity. When it enters an electric field that is perpendicular to its velocity, the electron will follow a ________ trajectory. When the electron enters a magnetic field that is perpendicular to its velocity, it will follow a ________ trajectory.

An alpha particle \(\left(m=6.6 \cdot 10^{-27} \mathrm{~kg}, q=+2 e\right)\) is accelerated by a potential difference of \(2700 \mathrm{~V}\) and moves in a plane perpendicular to a constant magnetic field of magnitude \(0.340 \mathrm{~T}\), which curves the trajectory of the alpha particle. Determine the radius of curvature and the period of revolution.

A particle with mass \(m\), charge \(q\), and velocity \(v\) enters a magnetic field of magnitude \(B\) and with direction perpendicular to the initial velocity of the particle. What is the work done by the magnetic field on the particle? How does this affect the particle's motion?
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