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Chapter 27: Problem 26

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?

Short Answer

Expert verified
The electron will have moved forward 1.16 * 10^-4 meters after completing one circle.

Step by step solution

01

Calculate the perpendicular component of the velocity

To find the perpendicular component of velocity, we'll use trigonometry and the given angle between the velocity and magnetic field lines (35°): v⊥ = v * sin(35°) Substitute the given value of electron speed: v⊥ = (4.0 * 10^5 m/s) * sin(35°) v⊥ ≈ 2.3 * 10^5 m/s
02

Calculate the radius of the helical path

Using the formula for helical path radius, we have: r = mv⊥/qB The mass of an electron (m) = 9.11 * 10^-31 kg The charge of an electron (q) = 1.6 * 10^-19 C Now plug in the values of v⊥, m, q, and B into the formula: r ≈ (9.11 * 10^-31 kg * 2.3 * 10^5 m/s) / (1.6 * 10^-19 C * 0.040 T) r ≈ 3.3 * 10^-4 m So, the radius of the helical path is 3.3 * 10^-4 meters.
03

Calculate the parallel component of the velocity

Similarly, we can find the parallel component of velocity by using the cosine of the given angle between the velocity and magnetic field lines (35°): v∥ = v * cos(35°) Substituting the given value of electron speed: v∥ = (4.0 * 10^5 m/s) * cos(35°) v∥ ≈ 3.3 * 10^5 m/s
04

Calculate the period of the circular motion

T = 2 * π * m / (q * B) Now plug in the values of m, q, and B into the formula: T ≈ 2 * π * (9.11 * 10^-31 kg) / (1.6 * 10^-19 C * 0.040 T) T ≈ 3.52 * 10^-10 s
05

Calculate the distance moved forward after completing one circle

Use the formula for the pitch of the helix: p = v∥T Now plug in the values of v∥ and T into the formula: p ≈ (3.3 * 10^5 m/s) * (3.52 * 10^-10 s) p ≈ 1.16 * 10^-4 m So, the electron will have moved forward 1.16 * 10^-4 meters after completing one circle. In conclusion, the radius of the helical path is 3.3 * 10^-4 meters, and the electron will have moved forward 1.16 * 10^-4 meters after completing one circle.

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Most popular questions from this chapter

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 straight wire of length \(2.00 \mathrm{~m}\) carries a current of \(24.0 \mathrm{~A} .\) It is placed on a horizontal tabletop in a uniform horizontal magnetic field. The wire makes an angle of \(30.0^{\circ}\) with the magnetic field lines. If the magnitude of the force on the wire is \(0.500 \mathrm{~N}\), what is the magnitude of the magnetic field?

Initially at rest, a small copper sphere with a mass of \(3.00 \cdot 10^{-6} \mathrm{~kg}\) and a charge of \(5.00 \cdot 10^{-4} \mathrm{C}\) is accelerated through a \(7000 .-\mathrm{V}\) potential difference before entering a magnetic field of magnitude \(4.00 \mathrm{~T}\), directed perpendicular to its velocity. What is the radius of curvature of the sphere's motion in the magnetic field?

In which direction does a magnetic force act on an electron that is moving in the positive \(x\) -direction in a magnetic field pointing in the positive \(z\) -direction? a) the positive \(y\) -direction b) the negative \(y\) -direction c) the negative \(x\) -direction d) any direction in the \(x y\) -plane

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?
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