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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

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 \(2 B\)

A copper wire with density \(\rho=8960 \mathrm{~kg} / \mathrm{m}^{3}\) is formed into a circular loop of radius \(50.0 \mathrm{~cm} .\) The cross-sectional area of the wire is \(1.00 \cdot 10^{-5} \mathrm{~m}^{2},\) and a potential difference of \(0.012 \mathrm{~V}\) is applied to the wire. What is the maximum angular acceleration of the loop when it is placed in a magnetic field of magnitude \(0.25 \mathrm{~T}\) ? The loop rotates about an axis through a diameter.

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