Explanation Given Info: The radius of the cylindrical wire is R . The current through the long cylindrical wire is I 0 . The one side length of the rectangle is W . Formula to calculate the magnetic field by Ampere’s law is, ∮ B → ⋅ d l → = μ 0 I encl (I) μ 0 is the magnetic constant, I encl is the current enclosed in the path, ∮ d l → is the differential length of the cylindrical wire. Formula to calculate the current density in the cylindrical wire is, J = I 0 A I 0 is the current flow in the cylindrical wire, A is the area of the cylindrical wire. Formula to calculate the area of the cylindrical wire is, A = π R 2 R is the radius of the cylindrical wire. Substitute π R 2 for A to find J . J = I 0 π R 2 (II) Refer figure.2. The current I encl enclosed in the cylindrical wire when r < R (inside the conductor) is, I encl = J A (III) Formula to calculate the area of the cylindrical wire when r < R is, A = π r 2 r is the distance from the center of the cylindrical wire. Substitute π r 2 for A in equation (III). I encl = J ( π r 2 ) (IV) Substitute equation (II) in equation (IV). I encl = ( I 0 π R 2 ) ( π r 2 ) = I 0 r 2 R 2 Substitute I 0 r 2 R 2 for I encl in equation (I). ∮ B → ⋅ d l → = μ 0 ( I 0 r 2 R 2 ) Take magnitude for the above equation, | ∮ B → ⋅ d l → | = | μ 0 ( I 0 r 2 R 2 ) | B ∮ d l = μ 0 I 0 r 2 R 2 (V) Substitute 2 π r for ∮ d l in equation (V). B ( 2 π r ) = μ 0 I 0 r 2 R 2 Rearrange the above equation as follows, B = μ 0 I 0 r 2 2 π r R 2 = μ 0 I 0 r 2 π R 2 (VI)

Mastering Physics with Pearson eText -- Standalone Access Card -- for University Physics with Modern Physics (14th Edition)

14th Edition

ISBN: 9780133978216

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Magnetic Force

A magnetic force arises when a charged particle or a current-carrying conductor is placed in a magnetic field. It is created because of the movement of the charged particles.

Question

Chapter 29, Problem 29.55P

To determine

The total magnetic flux through a rectangle.

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Chapter 29, Problem 29.54P

Chapter 29, Problem 29.56P

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Chapter 29 Solutions

Mastering Physics with Pearson eText -- Standalone Access Card -- for University Physics with Modern Physics (14th Edition)

Ch. 29.2 - The accompanying figure shows a wire coil being...Ch. 29.3 - (a) Suppose the magnet in Fig. 29.14a were...Ch. 29.4 - The earths magnetic field points toward (magnetic)...Ch. 29.5 - If you wiggle a magnet back and forth in your...Ch. 29.6 - Prob. 29.6TYU Ch. 29.7 - Prob. 29.7TYU Ch. 29 - A sheet of copper is placed between the poles of...Ch. 29 - Prob. 29.2DQ Ch. 29 - Prob. 29.3DQ Ch. 29 - Prob. 29.4DQ

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The total magnetic flux through a rectangle.

Solution Summary: The author explains how the magnetic flux through a rectangle is mu_0IW4pi.

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The total magnetic flux through a rectangle.

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Chapter 29 Solutions