Chapter 30

1 / 32

Chapter 30 - PowerPoint PPT Presentation

Chapter 30. Sources of Magnetic Field. Introduction. This chapter will focus on the sources of magnetic fields: moving charges. We’ll look at the field created by a current carrying conductor, as well as other symmetrical current distributions.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

PowerPoint Slideshow about 'Chapter 30' - nikki

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Chapter 30

Sources of Magnetic Field

Introduction
• This chapter will focus on the sources of magnetic fields: moving charges.
• We’ll look at the field created by a current carrying conductor, as well as other symmetrical current distributions.
• We’ll look at the force between two current-carrying conductors.
• We’ll finish by looking at the processes that result in materials being naturally magnetic.
30.1 The Biot-Savart Law
• Shortly after Oersted discovers that a compass needle is deflected by a nearby current-carrying conductor (1819) Jean-BaptisteBiot and Felix Savart begin quantitative experimentation.
• Their experimental results have given a mathematical expression for the magnetic field at some point in space in terms of the current that causes it.
30.1
• We will summarize results of their experiment concerning the magnetic field dB at a point P associated with a length element ds on a wire carrying current I.
30.1
• The results:
• The vector dB is perpendicular to both ds and r.
• The magnitude of dB is inversely proportional to r2, where r is the distance from ds to point P.
• The magnitude of dB is proportional to the current and to the magnitude of ds.
• The magnitude of dB is proportional to sin θ, where θ is the angle between ds and r.
30.1
• The mathematical expression summarizng these observations is known as the Biot-Savart Law:
• Remember μo is the permeability of free space
• Again, note that dB is only the field created by a single element of the conductor and to find the total magnetic field B at point P, we must integrate.
30.1
• Integration gives
• We must be careful with this integration as in does involved the vector cross product.
• For direction of the B field, the right hand rule is used.
• Point thumb in direction of I, curling fingers show the direction of B.
30.1
• Quick Quiz p. 928
• Example 30.1
• Resulting Equations 30.2, 30.3
30.2 Magnetic Force between Two Parallel Conductors
• If we have two current carrying wires, the B field caused by one current will exert a force on the other.
• If the conductors are parallel

to each other, then the B

field is perpendicular to the

current.

30.2
• The force on wire one from B-Field twois
• And assuming long wires ( << a)
• So the Force on wire one is given
30.2
• Often we make use of the Force per unit length
• By applying the right hand rule, we can see that two wires carrying current in the same direction will attract each other.
• Currents in the opposite direction will result in the conductors repelling.
30.2
• This is how the fundamental unit of the Ampere is defined.
• When the magnitude of the force per unit length between two long parallel wires that carry identical currents and are separated by 1 m is 2 x 10-7 N/m, the current in each wire is 1 A.
30.2
• From this the quantity of the Coulomb is defined.
• When a conductor carries a steady current of 1 A, the quantity of charge that flows through the cross section in 1 sec is 1 C.
• Quick Quizzes p. 933
30.2
• Ex: A long wire carries a current of 80 A. How much current must a second parallel wire carry if it is located 20 cm below the first wire such that it will not fall due to the force of gravity? Assume the lower wire has a linear density of 0.12 g/cm and a length of 1 m.
30.3 Ampere’s Law
• We know that a current will create a magnetic field in a circular path around a conductor.
30.3
• From symmetry we can assume that for a given circular path, where the conductor passes perpendicularly through the center, the magnitude of B is the same.
• We also know that B varies proportionally with current I, and inverse proportionally with the distance from it, a.
30.3
• If we look at the product of B and length element ds, and sum this around the circular path, this is called a line integral.
• B and ds are parallel to each other.
• And B is constant at radius r
30.3
• So the line integral goes as follows.
• But r and a are the same value, so the circumference cancels, giving Ampere’s Law
30.3
• Ampere’s Law- the line integral of B.ds around any closed path, equals μoI where I is the net steady current passing through any surface bound by the closed path.
• Ampere’s Law describes the creation of magnetic fields and will have similar application to Gauss’s Law, for applications of high symmetry.
30.3
• Quick Quizzes p. 934-935
• Example 30.4
30.3
• Example 30.5 “A Toroid”
30.3
• Example 30.6 “Infinite Current Sheet”
• Js is the linear current density

along the z axis in the picture

given.

30.4 Magnetic Field of Solenoid
• A solenoid is a long wire wound in a helix. It can create a reasonably uniform magnetic field in its interior.
• If the turns are tight together

and the solenoid has a finite

length, it closely resembles the

magnetic field of a bar magnet.

30.4
• By applying Ampere’s Law we find that the magnetic field inside a solenoid is given as
• Where N is the number of turns and L is the length of the solenoid, or n is the turns per unit length.
30.4
• Quick Quiz p. 940
30.5 Magnetic Flux
• Magnetic Flux is similar to Electric flux in that it describes the amount of electric field lines penetrating a surface.
• Consider and arbitrary object and element of surface area, dA.
• The flux through the

element is B.dA

30.5
• The total magnetic flux through the surface is is sum of the flux through each surface element.
• If the field is uniform at an angle θ, to the area vector then
30.5
• So if the field lines run parallel to the surface, θ = 90o and the flux is zero.
• If the field lines are perpendicular to the surface, then θ = 0o and the flux is a maximumn value.
30.5
• If the field is not uniform, and integration is often performed.
• Example 30.8 p. 941
30.6 Gauss’s Law for Magnetism
• Different from Gauss’s Law for Electric Fields.
• Electric Fields
• The net electric flux through a closed surface depends on the net charge inside (Qin)
• Magnetic Fields
• The net magnetic flux through any closed surface is zero.
• This is because all field lines are closed loops.
• They do not originate/terminate on discrete charges.