Explanation Given info: The length of the thin copper bar is 10.0 cm , the current in the copper bar in the negative x direction is 100 A , the distance between the bar and the straight wire is 0.500 cm and the current in the wire is 200 A in the positive z direction. The figure1 shows the direction of magnetic field in the copper bar. Figure (1) In the figure1 consider a small segment of length d x in the copper bar. The distance of the small segment from the left end of the copper bar is x . The magnetic field acts at angle θ with the vertical axis in the copper bar. From figure1 the expression of the magnetic force acts on the small segment is, d F = B cos θ × I 1 × d x (1) Here, B cos θ is the component of the magnetic field acts on the small segment. I 1 is the current in the copper bar. d x is the small segment in the copper bar. The expression of the magnetic field due to the straight wire is, B = μ 0 I 2 2 π ( x 2 + h 2 ) 1 2 Here, μ 0 is the permeability of free space. I 2 is the current in the straight wire. x is the distance of the small segment in the copper bar from the left end. h is the distance between the bar and the straight wire. Substitute μ 0 I 2 2 π ( x 2 + h 2 ) 1 2 for B in equation (1). d F = ( μ 0 I 2 2 π ( x 2 + h 2 ) 1 2 ) cos θ × I 1 × d x (2) The expression of cos θ from the figure1 is, cos θ = x ( x 2 + h 2 ) 1 2 Substitute x ( x 2 + h 2 ) 1 2 for cos θ in the equation (2). d F = ( μ 0 I 2 2 π ( x 2 + h 2 ) 1 2 ) ( x ( x 2 + h 2 ) 1 2 ) × I 1 × d x d F = ( μ 0 I 1 I 2 2 π ( x 2 + h 2 ) 1 2 ) ( x ( x 2 + h 2 ) 1 2 ) d x d F = μ 0 I 1 I 2 2 π ( x ( x 2 + h 2 ) ) d x Integrate the above equation over the whole length of the bar from the limit 0 to l

Physics for Scientists and Engineers, Volume 1, Chapters 1-22

8th Edition

ISBN: 9781439048382

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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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 30, Problem 30.71CP

To determine

The magnetic force exerted on the bar.

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Chapter 30 Solutions

Physics for Scientists and Engineers, Volume 1, Chapters 1-22

Ch. 30 - Consider the magnetic field due to the current in...Ch. 30 - A loose spiral spring carrying no current is hung...Ch. 30 - Prob. 30.3QQ Ch. 30 - Prob. 30.4QQ Ch. 30 - Consider a solenoid that is very long compared...Ch. 30 - Prob. 30.1OQ Ch. 30 - In Figure 30.7, assume I1 = 2.00 A ami I2 = 6.00...Ch. 30 - Answer each question yes or no. (a) Is it possible...Ch. 30 - Two long, parallel wires each carry the same...Ch. 30 - Two long, straight wires cross each other at a...

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