(a) Explanation Given info: The length of solenoid is l , radius is r , number of turns is N and steady current is I . Formula to calculate the magnetic field at a point due to a single turns of wire over the solenoid is, d B = μ 0 a 2 d I 2 ( a 2 + x 2 ) 3 / 2 (1) Here, μ 0 is the absolute permeability. a is the radius of the solenoid. x is the any point on the axis of the central line. Write the expression for the elementary current flow in a single turn of wire of width d r . d I = N I d r l Here, N is the total number of turns contains solenoid. I is the current flow in the wire. d r is the small width of the solenoid. l is the length of the solenoid. Replace d I by N I d r l in equation (1). d B = μ 0 a 2 2 ( a 2 + x 2 ) 3 / 2 ( N I d r l ) = μ 0 a 2 N I 2 l ( a 2 + x 2 ) 3 / 2 d r Write the expression for the total magnetic field produce by the solenoid. B = ∫ − x l − x d B Substitute μ 0 a 2 N I 2 l ( a 2 + x 2 ) 3 / 2 d r for d B (b) To determine To show: The magnetic field approaches μ 0 N I 2 l at each end of the solenoid as the length l becomes very long.

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN: 9781305116399

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Question

Chapter 30, Problem 30.69CP

(a)

To determine

The magnetic field at a point along the axis as a function of position x from the end of solenoid.

(b)

To determine

To show: The magnetic field approaches μ0NI2l at each end of the solenoid as the length l becomes very long.

Expert Solution & Answer

Chapter 30 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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 a point along the axis as a function of position x from the end of solenoid.

Solution Summary: The author explains the magnetic field at a point along the axis, based on the length of solenoid, radius, number of turns, and steady current.

Concept explainers

The magnetic field at a point along the axis as a function of position x from the end of solenoid.

To show: The magnetic field approaches μ0NI2l at each end of the solenoid as the length l becomes very long.

Chapter 30 Solutions