Magnetic Field problem in Griffith's book

In summary, the electromagnetic field is a way of representing the forces that charged particles exert on each other and the fields can vary greatly in both time and space.
  • #1
stunner5000pt
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4
Griffith's problem 6.14
For a bar magnet make careful sketches of M(Magnetization) , B (Magnetic Field),and H (Griffith's just calls this H...), assume L = 2a.

Ok since ther is no free current here
and [tex] \oint \vec{H} \bullet d \vec{l} = I_{f(enclosed)} [/tex]

H =0 yes??

The Attempt at a Solution


Check out my attached (bad) diagram. I drew the magnetic field. THe magnetization exists ONLY inside the magnet so it points in the direction from S to N. Since H = 0 it would not make an appearance... right?
 

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  • #3
OlderDan said:
What about B = μ(H + M)?

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/magpr.html

Try this out

http://www.walter-fendt.de/ph14e/mfbar.htm



hmm that formula does say that H would not be zero... is my magnetization right though? ZThere shouldn't be any outside the magnet...
 
  • #4
stunner5000pt said:
Griffith's problem 6.14
For a bar magnet make careful sketches of M(Magnetization) , B (Magnetic Field),and H (Griffith's just calls this H...), assume L = 2a.

Ok since ther is no free current here
and [tex] \oint \vec{H} \bullet d \vec{l} = I_{f(enclosed)} [/tex]

H =0 yes??


The Attempt at a Solution


Check out my attached (bad) diagram. I drew the magnetic field. THe magnetization exists ONLY inside the magnet so it points in the direction from S to N. Since H = 0 it would not make an appearance... right?
Wrong. H=B outside the magnet, but is in the opposite direction inside the magnet. H for the magnet is the same as E would be for two unilformly charged disks at the ends of the magnet. B is the same as for a solenoid.
 
  • #5
I didn't want to make a new topic for my question, so i put it here:

What is the (electro?)magnetic field? Could you say it is like an ocean, since I know that it is the reason for that electromagnetic waves can move, since the energy is "waves" like in the ocean. Is the field stronger some places than others? and does it have anything to do with the electric field that each particle has? (to bind themselves to each other, like electron (-) and proton (+))
 
  • #6
Jarle said:
I didn't want to make a new topic for my question, so i put it here:

What is the (electro?)magnetic field? Could you say it is like an ocean, since I know that it is the reason for that electromagnetic waves can move, since the energy is "waves" like in the ocean. Is the field stronger some places than others? and does it have anything to do with the electric field that each particle has? (to bind themselves to each other, like electron (-) and proton (+))

At the first level, an electric field is a way of representing the force one charged object exerts on another. This is the force that is responsible for binding electrons and protons together. A magnetic field is a way of representing an additional force that a moving charge exerts on another moving charge. When the fields associated with these forces are constant in time, we refer to these fields as being "static" fields. However, charges are often in motion in ways that cause both the fields they produce to change with time. Several great minds contributed to the development of a theory of these fields culminating in the work of Maxwell who showed that time varying electric and magnetic fields could propegate as "electromagnetic" waves in the empty space between charged particles.

As for the "ocean", physicists for a long time speculated about the existence of some ocean in empty space that was called the "ether". In the early 20th century, theoretical and experimental studies led to the conclusion that there is no such ocean needed for the electromagnetic wave to propegate through space.

The field varies a great deal in both time and space. The study of electromagnetic fields is a huge subject that has evolved from a classical wave perspective to a quantum electrodynamic perspective. There are several books dedicated to the subject, and numerous places on the web where you can find an introduction to the theory.
 

FAQ: Magnetic Field problem in Griffith's book

What is the Magnetic Field problem in Griffith's book?

The Magnetic Field problem in Griffith's book refers to a specific example problem in the field of electrodynamics that involves calculating the magnetic field produced by a moving charge. This problem is often used as an introductory exercise in understanding the principles of electromagnetic induction and the relationship between electricity and magnetism.

Why is the Magnetic Field problem important?

The Magnetic Field problem is important because it helps to illustrate the fundamental principles of electrodynamics and allows students to develop a deeper understanding of the relationship between electricity and magnetism. It also serves as a basis for more complex problems and applications in the field of electromagnetism.

How is the Magnetic Field problem solved?

The Magnetic Field problem is typically solved using vector calculus and the principles of electromagnetic theory. This involves applying Maxwell's equations and the Lorentz force law to the given scenario and using mathematical techniques to calculate the resulting magnetic field. The solution may also involve considering special cases and applying physical intuition to arrive at the correct answer.

What are some common challenges when solving the Magnetic Field problem?

Some common challenges when solving the Magnetic Field problem include understanding the underlying principles of electrodynamics, applying vector calculus and mathematical techniques correctly, and considering all relevant physical factors such as the direction and magnitude of the charge's velocity and the distance from the charge to the point where the magnetic field is being calculated.

How can I improve my understanding of the Magnetic Field problem?

To improve your understanding of the Magnetic Field problem, it is important to have a strong foundation in the principles of electrodynamics and vector calculus. You can also practice solving similar problems and seeking help from your peers or a teacher if you encounter any difficulties. Additionally, studying and understanding the derivations of the solution can also deepen your understanding of the problem and its applications.

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