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Murljashka [212]
3 years ago
5

What is the difference between the B-field and the H-field?

Physics
1 answer:
Simora [160]3 years ago
3 0
The H field is in units of amps/meter.  It is sometimes called the auxiliary field. It describes the strength (or intensity) of a magnetic field. The B field is the magnetic flux density.  It tells us how dense the field is.  If you think about a magnetic field as a collection of magnetic field lines, the B field tells us how closely they are spaced together. These lines (flux linkages) are measured in a unit called a Weber (Wb).  This is the analog to the electric charge, the Coulomb.  Just like electric flux density (the D field, given by D=εE) is Coulombs/m²,  The B field is given by Wb/m², or Tesla.  The B field is defined to be μH, in a similar way the D field is defined.  Thus B is material dependent.  If you expose a piece of iron (large μ) to an H field, the magnetic moments (atoms) inside will align in the field and amplify it.  This is why we use iron cores in electromagnets and transformers.
So if you need to measure how much flux goes through a loop, you need the flux density times the area of the loop Φ=BA.  The units work out like 
Φ=[Wb/m²][m²]=[Wb], which is really just the amount of flux.  The H field alone can't tell you this because without μ, we don't know the "number of field" lines that were caused in the material (even in vacuum) by that H field.  And the flux cares about the number of lines, not the field intensity.
I'm way into magnetic fields, my PhD research is in this area so I could go on forever.   I have included a picture that also shows M, the magnetization of a material along with H and B.  M is like the polarization vector, P, of dielectric materials. If you need more info let me know but I'll leave you alone for now!

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Answer:

heat used to rise temperature pan =  30.1%

heat used to rise temperature water =  69.9%

Explanation:

Given data

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solution

we find here heat transferred to the water that is

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here specific heat of aluminum is 900 J/kgºC

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heat transferred to the aluminum = 27 kJ

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total heat = 62.8 + 27 = 89.8 kJ

so

heat used to rise temperature pan = 27/89.8 ×100% = 30.1%

heat used to rise temperature water = 62.8 / 89.8 ×100% = 69.9%

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