Explanation Given info: The current in the straight wire I 1 is 5.00 A and current in the loop I 2 is 10.0 A . Distance between the loop and the straight wire c is 0.100 m , width of the loop a is 0.150 m and length of the loop l is 0.450 m . Consider the given figure as shown below. Figure (1) Write the expression for the magnetic force per unit length of the two parallel conductors. F B = μ 0 I 1 I 2 2 π x l Here, x is the distance between the wire. l is the length of the loop. μ 0 is the permeability of the free space. I 1 is the current in the straight wire. I 2 is the current in the circular loop. Write the expression for the magnetic force in the segment AB. F AB = μ 0 I 1 I 2 2 π c l Here, c is the distance between the two wire. l is the length of the loop. μ 0 is the permeability of the free space. I 1 is the current in the straight wire. I 2 is the current in the circular loop. Substitute 4 π × 10 − 7 T-m / A for μ 0 , 5.00 A for I 1 , 10.0 A for I 2 , 0.100 m for c and 0.45 m for l in the above expression for the magnetic force in the segment AB. F AB = ( 4 π × 10 − 7 T-m / A ) ( 5.00 A ) ( 10.0 A ) 2 π ( 0.100 m ) ( 0.45 m ) = 4.5 × 10 − 5 N Thus, the force on the wire AB is 4.5 × 10 − 5 N directed towards the wire. Write the expression for the magnetic force in the segment CD. F CD = μ 0 I 1 I 2 2 π ( c + a ) l Here, c + a is the distance between the two wire

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN: 9781305116399

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.

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Chapter 30, Problem 30.25P

To determine

The magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire.

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Chapter 30 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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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The magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire.

Solution Summary: The author explains the magnitude and direction of the net force exerted on the loop by the magnetic field created by a wire.

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The magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire.

Chapter 30 Solutions