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: 9781119306887

Author: Halliday

Publisher: WILEY

See similar textbooks

Concept explainers

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…

Question

Chapter 30, Problem 43P

To determine

To find:

Inductance per unit length of the wire.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 30, Problem 42P

Chapter 30, Problem 44P

Students have asked these similar questions

What is the force (in N) on the 2.0 μC charge placed at the center of the square shown below? (Express your answer in vector form.) 5.0 με 4.0 με 2.0 με + 1.0 m 1.0 m -40 με 2.0 μC

What is the force (in N) on the 5.4 µC charge shown below? (Express your answer in vector form.) −3.1 µC5.4 µC9.2 µC6.4 µC

An ideal gas in a sealed container starts out at a pressure of 8900 N/m2 and a volume of 5.7 m3. If the gas expands to a volume of 6.3 m3 while the pressure is held constant (still at 8900 N/m2), how much work is done by the gas? Give your answer as the number of Joules.

Chapter 30 Solutions

FUND.OF PHYSICS

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...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Inductance per unit length of the wire.

Concept explainers

To find: Inductance per unit length of the wire.

Want to see the full answer?

Chapter 30 Solutions

To find:

Inductance per unit length of the wire.