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 <

11th Edition

ISBN: 9781119459170

Author: Halliday

Publisher: WILEY

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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

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Ch. 30 - If the circular conductor in Fig. 30-21 undergoes...Ch. 30 - Prob. 2Q Ch. 30 - Prob. 3Q Ch. 30 - Prob. 4Q Ch. 30 - Prob. 5Q Ch. 30 - Prob. 6Q Ch. 30 - Prob. 7Q Ch. 30 - Prob. 8Q Ch. 30 - Prob. 9Q Ch. 30 - Prob. 10Q

Ch. 30 - Figure 30-31 shows three situations in which a...Ch. 30 - Figure 30-32 gives four situations in which we...Ch. 30 - Prob. 1P Ch. 30 - A certain elastic conducting material is stretched...Ch. 30 - Prob. 3P Ch. 30 - A wire loop of radius 12 cm and resistance 8.5 is...Ch. 30 - Prob. 5P Ch. 30 - Figure 30-37a shows a circuit consisting of an...Ch. 30 - In Fig. 30-38, the magnetic flux through the loop...Ch. 30 - Prob. 8P Ch. 30 - Prob. 9P Ch. 30 - Prob. 10P Ch. 30 - A rectangular coil of N turns and of length a and...Ch. 30 - Prob. 12P Ch. 30 - Prob. 13P Ch. 30 - GO In Fig. 30-42a, a uniform magnetic field B...Ch. 30 - GO A square wire loop with 2.00 m sides is...Ch. 30 - GO Figure 30-44a shows a wire that forms a...Ch. 30 - A small circular loop of area 2.00 cm2 is placed...Ch. 30 - Prob. 18P Ch. 30 - ILW An electric generator contains a coil of 100...Ch. 30 - At a certain place, Earths magnetic field has...Ch. 30 - Prob. 21P Ch. 30 - A rectangular loop area = 0.15 m2 turns in a...Ch. 30 - SSM Figure 30-47 shows two parallel loops of wire...Ch. 30 - Prob. 24P Ch. 30 - GO Two long, parallel copper wires of diameter 2.5...Ch. 30 - GO For the wire arrangement in Fig. 30-49, a =...Ch. 30 - ILW As seen in Fig. 30-50, a square loop of wire...Ch. 30 - Prob. 28P Ch. 30 - Prob. 29P Ch. 30 - Prob. 30P Ch. 30 - Prob. 31P Ch. 30 - A loop antenna of area 2.00 cm2 and resistance...Ch. 30 - GO Figure 30-54 shows a rod of length L = 10.0 cm...Ch. 30 - Prob. 34P Ch. 30 - Prob. 35P Ch. 30 - Prob. 36P Ch. 30 - Prob. 37P Ch. 30 - Prob. 38P Ch. 30 - Prob. 39P Ch. 30 - Prob. 40P Ch. 30 - A circular coil has a 10.0 cm radius and consists...Ch. 30 - Prob. 42P Ch. 30 - Prob. 43P Ch. 30 - Prob. 44P Ch. 30 - Prob. 45P Ch. 30 - Prob. 46P Ch. 30 - Inductors in series. Two inductors L1 and L2 are...Ch. 30 - Prob. 48P Ch. 30 - Prob. 49P Ch. 30 - Prob. 50P Ch. 30 - ILW The current in an RL circuit drops from 1.0 A...Ch. 30 - Prob. 52P Ch. 30 - Prob. 53P Ch. 30 - Prob. 54P Ch. 30 - Prob. 55P Ch. 30 - Prob. 56P Ch. 30 - In Fig. 30-65, R = 15 , L = 5.0 H, the ideal...Ch. 30 - Prob. 58P Ch. 30 - Prob. 59P Ch. 30 - Prob. 60P Ch. 30 - Prob. 61P Ch. 30 - A coil with an inductance of 2.0 H and a...Ch. 30 - Prob. 63P Ch. 30 - Prob. 64P Ch. 30 - Prob. 65P Ch. 30 - A circular loop of wire 50 mm in radius carries a...Ch. 30 - Prob. 67P Ch. 30 - Prob. 68P Ch. 30 - ILW What must be the magnitude of a uniform...Ch. 30 - Prob. 70P Ch. 30 - Prob. 71P Ch. 30 - Prob. 72P Ch. 30 - Prob. 73P Ch. 30 - Prob. 74P Ch. 30 - Prob. 75P Ch. 30 - Prob. 76P Ch. 30 - Prob. 77P Ch. 30 - Prob. 78P Ch. 30 - SSM In Fig. 30-71, the battery is ideal and = 10...Ch. 30 - Prob. 80P Ch. 30 - Prob. 81P Ch. 30 - A uniform magnetic field B is perpendicular to the...Ch. 30 - Prob. 83P Ch. 30 - Prob. 84P Ch. 30 - Prob. 85P Ch. 30 - Prob. 86P Ch. 30 - Prob. 87P Ch. 30 - Prob. 88P Ch. 30 - A coil with an inductance of 2.0 H and a...Ch. 30 - Prob. 90P Ch. 30 - Prob. 91P Ch. 30 - Prob. 92P Ch. 30 - Prob. 93P Ch. 30 - A long cylindrical solenoid with 100 turns/cm has...Ch. 30 - Prob. 95P Ch. 30 - A square loop of wire is held in a uniform 0.24 T...Ch. 30 - Prob. 97P Ch. 30 - The inductance of a closely wound coil is such...Ch. 30 - The magnetic field in the interstellar space of...Ch. 30 - Prob. 100P Ch. 30 - A toroid has a 5.00 cm square cross section, an...

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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.