Explanation Given data: The center of the loop is at the origin and its long sides parallel to the x axis. The current in the loop is 50 A . Calculation: A rectangular loop of wire situated in x-y plane with the center of the loop at the origin is shown in Figure 1. It is observed that the field from each of the segments adds up at the center and the direction of the magnetic field changes but the magnitude of the magnetic field is the same. The direction of the magnetic field in segment 1 is calculated as, − y ^ × ( − x ^ ± y ^ ) = − z ^ The direction of the magnetic field in segment 2 is calculated as, − x ^ × ( ± x ^ − y ^ ) = − z ^ The direction of the magnetic field in segment 3 is calculated as, y ^ × ( x ^ ± y ^ ) = − z ^ The direction of the magnetic field in segment 4 is calculated as, − x ^ × ( ± x ^ + y ^ ) = − z ^ The direction of the magnetic field in each segment is − z ^ . The expression for the magnetic field at a point due to a linear each segment is given as, B = − z ^ ( μ 0 I l 2 π r 4 r 2 + l 2 ) (1) Here, B is the magnetic field at a point due to a linear each segment, μ 0 is the magnetic permeability ( μ 0 = 4 π × 10 − 7 ) , I is the current, l is the length, r is the distance of the point, The conversion of cm into m is done as, 1 cm = 1 × 10 − 2 m Hence, the conversion of 8 cm into m is, 8 cm = 8 × 10 − 2 m Hence, the conversion of 6 cm into m is, 6 cm = 6 × 10 − 2 m The magnetic field due to segment 1 at the center of the loop is calculated as, B 1 = − z ^ ( μ 0 I l 2 π r 4 r 2 + l 2 ) Substitute 8 × 10 − 2 for l , 6 × 10 − 2 for r and 50 for I in the above formula. B 1 = − z ^ ( 4 π × 10 − 7 × 50 × 8 × 10 − 2 2 π × 6 × 10 − 2 4 ( 6 × 10 − 2 ) 2 + ( 8 × 10 − 2 ) 2 ) = − z ^ ( 133.33 × 10 − 7 4 ( 6 × 10 − 2 ) 2 + ( 8 × 10 − 2 ) 2 ) = − z ^ ( 9.24 × 10 − 5 ) T The magnetic field due to segment 2 at the center of the loop is calculated as, B 2 = − z ^ ( μ 0 I l 2 π r 4 r 2 + l 2 ) Substitute 12 × 10 − 2 for l , 4 × 10 − 2 for r and 50 for I in the above formula. B 2 = − z ^ ( 4 π × 10 − 7 × 50 × 12 × 10 − 2 2 π × 4 × 10 − 2 4 ( 4 × 10 − 2 ) 2 + ( 12 × 10 − 2 ) 2 ) = − z ^ ( 300 × 10 − 7 4 ( 4 × 10 − 2 ) 2 + ( 12 × 10 − 2 ) 2 ) = − z ^ ( 20.80 × 10 − 5 ) T The magnetic field due to segment 3 at the center of the loop is calculated as, B 3 = − z ^ ( μ 0 I l 2 π r 4 r 2 + l 2 ) Substitute 8 × 10 − 2 for l , 6 × 10 − 2 for r and 50 for I in the above formula

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ISBN: 8220100663659

Author: ULABY

Publisher: PEARSON

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Chapter 5, Problem 7P

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The magnetic flux density at the center of the loop.

Solution Summary: The center of the loop is at the origin and its long sides parallel to the x axis.

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The magnetic flux density at the center of the loop.

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