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!
Answer:
600m
Explanation:
30×20 at a constant speed is 600m.
Answer:
0.5m
Explanation:
v=f×lamda
v is 300m/s, f is 600Hz, lamda is ?
lamda=v/f
lamda=300/600
lamda =3/6=1/2m
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q = 
=
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is
.
To stop instantly, you would need infinite deceleration. This in turn, requires infinite force, as demonstrable with this equation:F=ma<span>So when you hit a wall, you do not instantly stop (e.g. the trunk of the car will still move because the car is getting crushed). In a case of a change in momentum, </span><span><span>m<span>v⃗ </span></span><span>m<span>v→</span></span></span>, we can use the following equation to calculate force:F=p/h<span>However, because the force is nowhere close to infinity, time will never tend to zero either, which means that you cannot come to an instantaneous stop.</span>