To determine The magnitude of the current in wire 1 and its direction. Explanation It is given that magnitudes of the currents in both wires are same. Take this current to be I . Write the equation for the magnetic field due to the wire 1 carrying current. B 1 = μ 0 I 2 π r (I) Here, B 1 is the magnetic field due to wire 1 at perpendicular distance r from the wire, μ 0 is the absolute permeability, I is the current and r is the perpendicular distance between the point P and wire 1 Write the equation for the magnetic field due to the wire 2 carrying current. B 2 = μ 0 I 2 π ( 3 r ) = μ 0 I 6 π r (II) Here, B 2 is the magnetic field due to wire 2 at perpendicular distance 3 r from the wire and ( 3 r ) is the perpendicular distance between the point P and wire 2 Comparison of equations (I) and (II) shows that B 2 < B 1 so that the direction of the net magnetic field at point P is the same as the direction of the field due to wire 1. According to right hand rule, wire 1’s current must be into the page, since magnetic field at P due to wire 1 acts in downward direction.. Since the currents in the wires are in opposite directions, wire 2’s current is out of the page. Both the fields act downward so that the net field at P also acts downward. Write the equation for the net field at point P . B → net = ( B 1 + B 2 ) down Here, B → net is the net magnetic field at point P Put equations (I) and (II) in the above equation and rewrite the equation for I . B → net = μ 0 I 2 π r + μ 0 I 6 π r = 2 μ 0 I 3 π r I = 3 π r B net 2 μ 0 (III) Conclusion: Given that the value of r is 2

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ISBN: 9781260486919

Author: GIAMBATTISTA

Publisher: MCG

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Chapter 19, Problem 78P

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The magnitude of the current in wire 1 and its direction.

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Chapter 19, Problem 77P

Chapter 19, Problem 79P

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