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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 rail gun accelerates a projectile from rest by using the magnetic force on a current-carrying wire. The wire has radius \(r=5.1 \cdot 10^{-4} \mathrm{~m}\) and is made of copper having a density of \(\rho=8960 \mathrm{~kg} / \mathrm{m}^{3}\). The gun consists of rails of length \(L=1.0 \mathrm{~m}\) in a constant magnetic field of magnitude \(B=2.0 \mathrm{~T}\) oriented perpendicular to the plane defined by the rails. The wire forms an electrical connection across the rails at one end of the rails. When triggered, a current of \(1.00 \cdot 10^{4}\) A flows through the wire, which accelerates the wire along the rails. Calculate the final speed of the wire as it leaves the rails. (Neglect friction.)

The work done by the magnetic field on a charged particle in motion in a cyclotron is zero. How, then, can a cyclotron be used as a particle accelerator, and what essential feature of the particle's motion makes it possible? A blue problem number indicates a worked-out solution is available in the Student Solutions Manual. One \(\bullet\) and two \(\bullet\) indicate increasing level of problem difficulty.

It would be mathematically possible, for a region with zero current density, to define a scalar magnetic potential analogous to the electrostatic potential: \(V_{B}(\vec{r})=-\int_{\vec{r}_{0}}^{\vec{r}} \vec{B} \cdot d \vec{s},\) or \(\vec{B}(\vec{r})=-\nabla V_{B}(\vec{r}) .\) However, this has not been done. Explain why not.

An electron moves in a circular trajectory with radius \(r_{\mathrm{i}}\) in a constant magnetic field. What is the final radius of the trajectory when the magnetic field is doubled? a) \(\frac{r_{i}}{4}\) b) \(\frac{r_{i}}{2}\) c) \(r_{i}\) d) \(2 r_{i}\) e) \(4 r_{\mathrm{i}}\)

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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