Solenoids and Force in Different Situations

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The discussion focuses on the behavior of solenoids and magnetic fields, particularly regarding the forces experienced by a metal bar and a bar magnet when interacting with a solenoid's magnetic field. When a metal bar is inserted into the solenoid, it experiences a force due to the magnetic field, which is directed towards the solenoid. Inserting a bar magnet into the solenoid, whether North or South end first, results in attraction if the magnet is made of ferromagnetic material, as it seeks a lower energy state. The interaction between the solenoid's magnetic field and the magnet behaves similarly to two magnets, with attraction in one orientation and repulsion in the other. The discussion also emphasizes the importance of understanding magnetic field direction and the calculations involved in determining electromagnetic force.
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I'm currently studying about Solenoids and Magnetic Fields. However, some questions are confusing me and I'm not really sure how to approach them. If I can get any input on them, I'd really appreciate it. ps, these questions aren't homework questions, they're questions that haven't been assigned since they're at a higher difficulty and I'm curious as to how they work out.

A solenoid is connected to a DC power supply such that the magnetic field is pointing out of the solenoid's on the end that is facing you.

a) a metal bar about 50% longer than the solenoid is inserted halfway into the solenoid. Does the metal bar experience any force from the solenoid? If so, in what direction?

b) i) With the metal bar removed, if you were to attempt to insert a bar magnet from the solenoid side where the magnetic field is going outward, North end first, what force, if any, would it experience?
ii) What if the bar magnet is inserted South end first?
 
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In my opinion the bar-if it is a ferromagnetic metal made-will be attracted into the solenoid, it does not matter how will be introduced in the solenoid. The magnetic flow creates magnetic moments in the bar which represents a kind of magnet and will react with the flow-as in a lifting electromagnet.
The force is actually the lifting electromagnet force:
F=B^2*A:8\pi:9.81.10^5 =4.05/10^8*B^2*A kgf B [Wb/m^2] A[m^2] solenoid cross section area.
B=\mu_o.H
H=N*I/h [A/m]
\mu_o=4\pi:10^7 [H/m] N=solenoid no. of turns I[A] current h[m] solenoid length.
 
Notes on your questions:
The "direction" of the magnetic field is arbitrarily assigned by us. The Field is always in a continuous loop- We refer to these as North and South - because the first magnetic field recognized as that of the Earth, and it aligned the north direction. ( It is interesting to note - in our naming convention the North pole of the Earth - would actually be labeled as the S pole if it were a magnet! That is why magnets N is attracted to the North pole of the earth)

The fero-magnetic material is "pulled" into the field, reducing the size of the magnetic field overall. The force is generated as the "system" seeks a lower energy state. It is easier to create the M field in the iron, than in the air - the iron supports higher flux density- the same amount of Flux, in less area,,, lower energy.

As for a MAGNETIC material - the field of the magnet and the solenoid interact exactly the same as two magnets. In one polarity they attract and in the other they repel. Reversing the direction of the current in the solenoid - has the same effect as turning the magnet around. IN one orientation they attract, in the other they repel.
 
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