Explanation 1) Concept: The current flowing through the loop ABCDEFA will produce a magnetic dipole moment. But to determine its value, we need to breakdown the loop into two smaller loops, ABCDA and ADEFA. We can calculate their magnetic dipole moments independently and perform the vector addition of the two. 2) Formula: μ = N i A 3) Given: i) The current through the loop = i = 5.0 A ii) The sides of the loop are parallel to the coordinate axes as shown in the figure. iii) The length of the sides, A B = 20.0 c m = 0.20 m , B C = 30.0 c m = 0.30 m and F A = 10.0 c m = 0.10 m 4) Calculations: The loop ABCDEFA is broken down into two smaller loops namely, ABCDA and ADEFA. We imagine equal and opposite currents i in the imaginary segment AD. The magnetic moment of the loop ABCDA is calculated as μ = N i A N = 1 and A = A r e a o f r e c t a n g l e A B C D ∴ μ 1 = i × ( A B × B C ) μ 1 = 5.0 × 0.20 × 0.30 μ 1 = 0.30 A . m 2 The direction of the magnetic moment is given by the right-hand rule. By applying the rule, we find that the direction is along – Z axis since the current is flowing clockwise. ∴ μ 1 = - 0

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Author: David Halliday, Robert Resnick, Jearl Walker

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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 28, Problem 60P

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The magnetic moment of the loop in unit vector notation

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Chapter 28, Problem 59P

Chapter 28, Problem 61P

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The magnetic moment of the loop in unit vector notation

Solution Summary: The author explains how the current flowing through the loop ABCDEFA will produce a magnetic dipole moment, but to determine its value, we need to break it into two smaller loops, ABCDA and ADEFA.

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To find: The magnetic moment of the loop in unit vector notation

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Chapter 28 Solutions

To find:

The magnetic moment of the loop in unit vector notation