(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.

9th Edition

ISBN: 9781305804470

Author: Jewett

Publisher: CENGAGE LEARNING - CONSIGNMENT

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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.

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Chapter 30 Solutions

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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