Explanation 1) Concept: The crucial step in this problem is to find the magnetic flux due to current- carrying wires. Using the flux we can find inductance of the given system; dividing inductance by unit length, we will get the required answer 2) Formula: i) Magnitude of magnetic field at radial distance r from the center of the long wire (inside) B = μ 0 i 2 π · r a 2 ii) Magnitude of magnetic field at radial distance r from the center of the long wire (inside) B = μ 0 i 2 π r iii) Magnetic flux, ϕ B = L i iv) Magnetic flux due to B field across open surface attached to the closed loop is ϕ B = ∮ B → · d A → 3) Given: i) Radius of each long wire, a = 1.53 × 10 - 3 m = 0.153 × 10 - 2 m ii) Center to center separation between two wires d = 14.2 c m = 14.2 × 10 - 2 m iii) Unit length of wire, l = 1.0 m 4) Calculations: Let’s denote two wires as w 1 and w 2 Now we will find B field due to wire w 1 and w 2 at point p inside w 1 at distance r from the center of w 1 which is calculated by using amperes law B i n = μ 0 i 2 π · r a 2 + μ 0 i 2 π ( d - r ) In the above expression first term is due to w 1 and second term is due to w 2 Similarly we will find B field due to wire w 1 and w 2 at point p outside, at distance r from the center of w 1 which is calculated by using amperes law B o u t = μ 0 i 2 π r + μ 0 i 2 π ( d - r ) In the above expression first term is due to w 1 and second term is due to w 2 Total field in the region between w 1 and w 2 is then B i n + B o u t = μ 0 i 2 π · r a 2 + μ 0 i 2 π d - r + μ 0 i 2 π r + μ 0 i 2 π d - r Magnetic flux due to B field is ϕ B = ϕ B 1 + ϕ B 2 = ∮ B → · d A → = ∫ 0 d / 2 B · d A + ∫ d / 2 d B · d A = 2 ∫ 0 d / 2 B · d A ϕ B = 2 ∫ 0 a B i n · d A + 2 ∫ a d 2 B o u t · d A <

10th Edition

ISBN: 9781118230725

Author: David Halliday, Robert Resnick, Jearl Walker

Publisher: Wiley, John & Sons, Incorporated

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Voltage Across Inductor

An actor is a passive electrical component that consists of a coil bent in the form of wire which uses electromagnetism to produce electric current through the coil. When the electric current flows through a conductor that is the coil, a magnetic flux is…

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Chapter 30, Problem 43P

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Inductance per unit length of the wire.

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Chapter 30, Problem 42P

Chapter 30, Problem 44P

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Chapter 30 Solutions

Fundamentals of Physics Extended

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Inductance per unit length of the wire.

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To find: Inductance per unit length of the wire.

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Chapter 30 Solutions

To find:

Inductance per unit length of the wire.