Flux Density due to a current flowing in a long straight wire

Slides:



Advertisements
Similar presentations
Magnetic Forces and Fields
Advertisements

Gauss’ Law AP Physics C.
Topic 6.3: Magnetic force and field
Electromagnetic Induction Inductors. Problem A metal rod of length L and mass m is free to slide, without friction, on two parallel metal tracks. The.
Copyright © 2009 Pearson Education, Inc. Chapter 21 Electric Charge and Electric Field.
Measuring the strength of a Magnetic Field © David Hoult 2009.
Topic 12.1 Induced electromotive force (emf) 3 hours.
Example: A single square loop of wire with a resistance of 100  and 10 cm long sides is rotated from 30 o to 60 o in a constant magnetic field of 5T.
Chapter 28 Sources of Magnetic Field
Induced EMF and Inductance 1830s Michael Faraday Joseph Henry M is mutual inductance.
Current, Ohm’s Law, Etc. The Continuity Equation for Steady State Currents Currents and current densities are constant in time – steady state. The flux.
III–2 Magnetic Fields Due to Currents.
I-2 Gauss’ Law Main Topics The Electric Flux. The Gauss’ Law. The Charge Density. Use the G. L. to calculate the field of a.
General Physics 2Magnetism1 E9 – Currents Create Magnetic Fields A moving charge creates a magnetic field Superposition principle for a wire segment Biot-Savart.
Nadiah Alanazi Gauss’s Law 24.3 Application of Gauss’s Law to Various Charge Distributions.
Two questions: (1) How to find the force, F on the electric charge, q excreted by the field E and/or B? (2) How fields E and/or B can be created?
By: Engr. Hinesh Kumar (Lecturer) MAGNETIC FLUX & MAGNETIC FLUX DENSITY.
Electricity and Magnetism: Electromagnets Mr D. Patterson.
Inductors Chap 11.
Electro-Magnetic Induction © David Hoult Magnetic flux © David Hoult 2009.
Flux Density due to a current flowing in a long straight wire © David Hoult 2009.
Electro-Magnetism © David Hoult Magnetic Field Shapes © David Hoult 2009.
Chapter 20 The Production and Properties of Magnetic Fields.
Gauss’ Law for magnetic fields There are no magnetic “charges”.
1 Magnetostatics. 2 If charges are moving with constant velocity, a static magnetic (or magnetostatic) field is produced. Thus, magnetostatic fields originate.
EEL 3472 Magnetostatics 1. If charges are moving with constant velocity, a static magnetic (or magnetostatic) field is produced. Thus, magnetostatic fields.
Physics 2102 Magnetic fields produced by currents Physics 2102 Gabriela González.
Magnetic Fields in Wires. Strength of Magnetic Field Strength of the magnetic field produced by a current carrying wire is directly proportional to the.
1 Propagation of waves Friday October 18, Propagation of waves in 3D Imagine a disturbane that results in waves propagating equally in all directions.
Chapter E9 Symmetry and Flux E9B.1, E9B.2, E9B.5, E9B.9 and E9S.6 (only part a). Due Wednesday.
Copyright © 2009 Pearson Education, Inc. Chapter 22 Gauss’s Law.
Two questions: (1) How to find the force, F on the electric charge, Q excreted by the field E and/or B? (2) How fields E and/or B can be created?
Chapter 26 Sources of Magnetic Field. Biot-Savart Law (P 614 ) 2 Magnetic equivalent to C’s law by Biot & Savart . P. P Magnetic field due to an infinitesimal.
Gauss’s law This lecture was given mostly on the board so these slides are only a guide to what was done.
More Examples of Magnetic Flux Biot-Savart Law Ampère’s Law.
Wave Equations: EM Waves. Electromagnetic waves for E field for B field.
Week 9 Presentation 1 Electromagnets 1. Learning Objectives: 1. Determine the magnitude and direction of the magnetic field strength generated by a straight.
Magnetism. Magnets and Magnetic Fields Magnets have two ends – poles – called north and south. Like poles repel; unlike poles attract.
3/21/20161 ELECTRICITY AND MAGNETISM Phy 220 Chapter2: Gauss’s Law.
1 15. Magnetic field Historical observations indicated that certain materials attract small pieces of iron. In 1820 H. Oersted discovered that a compass.
1© Manhattan Press (H.K.) Ltd magnetic fields due to currents Long straight wire Circular coil Long solenoid.
Magnetic Fields Starter questions
The Biot-Savart Law. Biot and Savart recognized that a conductor carrying a steady current produces a force on a magnet. Biot and Savart produced an equation.
Magnets have two ends – poles – called north and south. Like poles repel; unlike poles attract. If you cut a magnet in half, you don’t get a north pole.
Two questions: (1) How to find the force, F on the electric charge, q excreted by the field E and/or B? (2) How fields E and/or B can be created?
24.2 Gauss’s Law.
2.2 ELECTROMAGNETISM 19th November 2012
y P  dB r a  x  z ds I x When L, or
Electro-Magnetism © D Hoult 2008.
Chapter 3 Magnetostatics
Force acting on a charged particle moving through a magnetic field
Physics 014 Gauss’ Law.
Magnetic Field produced by current
Electro-Magnetic Induction
Magnetism.
Example: calculate the magnetic field at point P due to a thin straight wire of length L carrying a current I. (P is on the perpendicular bisector of the.
Parallel Currents d L Current I1 produces a B-field I1 I2 B1 xxxx
Two questions: (1) How to find the force, F on the electric charge, q excreted by the field E and/or B? (2) How fields E and/or B can be created?
Flux density produced by a long coil (solenoid)
Measuring the strength of a Magnetic Field
Force acting between two long, parallel, current-carrying conductors
Chapter 22 Gauss’s Law Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s.
Gauss’s law This lecture was given mostly on the board so these slides are only a guide to what was done.
The Motion of Charged Particles in Magnetic Fields
Sources of Magnetic Fields
Chapter 22 Gauss’s Law Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s.
Applying Gauss’s Law Gauss’s law is useful only when the electric field is constant on a given surface 1. Select Gauss surface In this case a cylindrical.
Example: calculate the magnetic field at point P due to a thin straight wire of length L carrying a current I. (P is on the perpendicular bisector of the.
Chapter 30 Examples 4,8.
Presentation transcript:

Flux Density due to a current flowing in a long straight wire © D Hoult 2008

The field at point p is directed

The field at point p is directed out of the plane of the diagram (“corkscrew rule”)

The magnitude of B at point p depends on

The magnitude of B at point p depends on the current, I

The magnitude of B at point p depends on the current, I the perpendicular distance of p from the wire

The magnitude of B at point p depends on the current, I the perpendicular distance of p from the wire the medium surrounding the wire

Experiments show that B a I and if r is small compared with the length of the wire then

Experiments show that B a I and if r is small compared with the length of the wire then 1 B a r Therefore

Experiments show that B a I and if r is small compared with the length of the wire then 1 B a r Therefore I B = (a constant) r

Because this is a situation having cylindrical symmetry, the factor 2p is included in the equation

Because this is a situation having cylindrical symmetry, the factor 2p is included in the equation 2 p r

Because this is a situation having cylindrical symmetry, the factor 2p is included in the equation 2 p r where µ is the permeability of the medium surrounding the wire

Because this is a situation having cylindrical symmetry, the factor 2p is included in the equation 2 p r where µ is the permeability of the medium surrounding the wire If the medium is a vacuum (or air) the permeability is written as µo

Because this is a situation having cylindrical symmetry, the factor 2p is included in the equation 2 p r where µ is the permeability of the medium surrounding the wire If the medium is a vacuum (or air) the permeability is written as µo The units of µ are

Because this is a situation having cylindrical symmetry, the factor 2p is included in the equation 2 p r where µ is the permeability of the medium surrounding the wire If the medium is a vacuum (or air) the permeability is written as µo The units of µ are T m A-1 =

Because this is a situation having cylindrical symmetry, the factor 2p is included in the equation 2 p r where µ is the permeability of the medium surrounding the wire If the medium is a vacuum (or air) the permeability is written as µo The units of µ are T m A-1 = NA-2

Because this is a situation having cylindrical symmetry, the factor 2p is included in the equation 2 p r where µ is the permeability of the medium surrounding the wire If the medium is a vacuum (or air) the permeability is written as µo The units of µ are T m A-1 = NA-2 1 N A-2 = 1 Henry per meter (H m-1)