Explanation 1) Concept: By using the formula for kinetic energy and magnetic force acting on the moving charged particle, we can find the value of acceleration and distance of the electron. 2) Formula: i) Magnetic force, F B → = q v → × B → ii) Kinetic energy of particle, K = 1 2 m v 2 3) Given: i) Kinetic energy of electron K = 12 k e V = 12 × 10 3 × 1.6 × 10 - 19 = 19.2 × 10 - 16 J ∴ K = 19.2 × 10 - 16 J ii) Vertical component of earth’s magnetic field B = 55 μ T = 55 × 10 - 6 T 4) Calculations: a) The net force acting on an electron is the vertical component of earth’s magnetic field, which is perpendicular to the motion of the electron. F B = F n e t = m a Using F n e t = F B → = q v → × B → F n e t = q v B sin ⁡ φ As the direction of magnetic field is perpendicular to the velocity of the electron, φ = 90 0 , sin ⁡ 90 = 1 F n e t = q v B ∴ m a = q v B a = q v B m … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … 1 Kinetic energy of the particle, K = 1 2 m v 2 v = 2 K m Using in equation 1, we get a = q B m 2 K m a = q B 2 K m 3 For the electron, q = 1.6 × 10 - 19 a n d m = 9.11 × 10 - 31 k g , and using all the given values, a = 1

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Author: David Halliday, Robert Resnick, Jearl Walker

Publisher: Wiley, John & Sons, Incorporated

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Magnetic Force

A magnetic force arises when a charged particle or a current-carrying conductor is placed in a magnetic field. It is created because of the movement of the charged particles.

Question

Chapter 28, Problem 73P

To determine

To find:

a) Magnitude of the electron’s acceleration (a)

b) Electron’s distance from x axis when x = 20 cm

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Chapter 28, Problem 72P

Chapter 28, Problem 74P

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Chapter 28 Solutions

Fundamentals of Physics Extended

Ch. 28 - Prob. 1Q Ch. 28 - Prob. 2Q Ch. 28 - Prob. 3Q Ch. 28 - Prob. 4Q Ch. 28 - In Module 28-2, we discussed a charged particle...Ch. 28 - Figure 28-26 shows crossed uniform electic and...Ch. 28 - Figure 28-27 shows the path of an electron that...Ch. 28 - Figure 28-28 shows the path of an electron in a...Ch. 28 - a In Checkpoint 5, if the dipole moment is rotated...Ch. 28 - Particle round about. Figure 28-29 shows 11 paths...

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a) Magnitude of the electron’s acceleration (a) b) Electron’s distance from x axis when x = 20 cm

Solution Summary: The author explains how the formula for kinetic energy and magnetic force acting on the moving charged particle can find the value of acceleration and distance of the electron.

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To find: a) Magnitude of the electron’s acceleration (a) b) Electron’s distance from x axis when x = 20 cm

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Chapter 28 Solutions

To find:

a) Magnitude of the electron’s acceleration (a)

b) Electron’s distance from x axis when x = 20 cm