Explanation Formula to calculate the magnetic field at a point due to straight wire segment is, B = μ 0 I 4 π R ( 2 sin θ ) (1) Here, B is the magnetic field. μ 0 is the absolute permittivity. I is the current flows in the straight wire. x is the distance of the of point from the center of the plane. θ is the angle made by the point to the edge of the wire. Write the expression for the distance of the point from the x − y plane is, R = x 2 + ( L / 2 ) 2 = x 2 + L 4 2 Write the expression for the distance of the point from the x − y plane is, d = x 2 + R 2 Replace R 2 by x 2 + L 4 2 in above equation. d = L 4 2 + x 2 + L 4 2 = x 2 + L 2 2 Write the expression for the sine of the angle of sin θ = L 2 d Substitute x 2 + L 2 2 for d in above equation. sin θ = L 2 x 2 + L 2 2 = L 2 x 2 + L 2 2 Substitute x 2 + L 4 2 for R and L 2 x 2 + L 2 2 for sin θ in equation (1). B = μ 0 I 4 π ( x 2 + L 4 2 ) ( 2 L 2 x 2 + L 2 2 ) = μ 0 I L 4 π ( x 2 + L 4 2 ) ( x 2 + L 2 2 ) Since the shape of the wire is symmetrical, hence all the magnetic field in the y and z direction will be cancel out to each other

Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term

9th Edition

ISBN: 9781305714892

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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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 30, Problem 30.76CP

To determine

To show: The magnetic field at point P a distance x from the center of the square along its axis is B=μ0IL22π(x2+L2/4)x2+L2/2 .

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Chapter 30 Solutions

Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term

Ch. 30 - Consider the magnetic field due to the current in...Ch. 30 - A loose spiral spring carrying no current is hung...Ch. 30 - Prob. 30.3QQ Ch. 30 - Prob. 30.4QQ Ch. 30 - Consider a solenoid that is very long compared...Ch. 30 - Prob. 30.1OQ Ch. 30 - In Figure 30.7, assume I1 = 2.00 A ami I2 = 6.00...Ch. 30 - Answer each question yes or no. (a) Is it possible...Ch. 30 - Two long, parallel wires each carry the same...Ch. 30 - Two long, straight wires cross each other at a...

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The magnetic field at point P a distance x from the center of the square along its axis is B = μ 0 I L 2 2 π ( x 2 + L 2 / 4 ) x 2 + L 2 / 2 .

Solution Summary: The author explains the formula to calculate the magnetic field at a point due to straight wire segment.

Concept explainers

To show: The magnetic field at point P a distance x from the center of the square along its axis is B=μ0IL22π(x2+L2/4)x2+L2/2 .

Chapter 30 Solutions