# Tag Info

5

"Magnetic force" does not need to do work for the "magnetic energy" to change. There is no work-energy relation between these two concepts only. The magnetic field is said to store magnetic energy because to create it, work has to be done by external bodies (against induced electrical forces originating in the system), and this work can be recovered if the ...

2

Note that the voltage induced by the changing magnetic field is directional. To reduce the resulting currents, you only need to increase resistivity in the direction the current would flow. That's what laminations do. Laminations are thin sheets of metal that conduct electricity (as a unintentional side effect of having desirable magnetic properties). ...

1

First, your force equation is wrong, as you're missing the electric field. Wait what electric field? That's the point! A changing magnetic field induces an electric field $\nabla\times E=-\frac{\partial B}{\partial t}$, and this "pushes" the current. Note that the applied magnetic field is perpendicular to the circuit/wire, so that at least part of the ...

1

the angle between $dl$ and $r$ is $\pi/2$, which is not $\theta$. $\theta$ is the angle between $r$ and y-z plane. if you know what is $x$ and $r$, then $\sin\theta=x/r$, where $x$ is the distance from the origin to the point of intersection of $r$ and x-axis, and $r$ is the distance from p point to the same place UPDATE. $\theta$ is important because your ...

1

Line integrals of the magnetic field strength are magnetic voltage drops. Just google for "magnetic voltage drop" (including the double-quotes). In the quasi-static case ($\dot{\vec{D}}=\vec{0}$) the $\vec{H}$-field within a simply path-connected domain with zero current density has a magnetic potential. In this case you can calculate the magnetic voltage ...

1

Firstly, if your waveguide is a hollow conductor, it cannot support TEM modes. There must be at least two separate (electrically insulated from one another) conductors in the waveguide's cross section for TEM modes to propagate. The reason is that the transverse field dependence of a TEM mode is the same as that of a static field, as I explain in detail in ...

1

The circulation is indeed zero. You are confusing flux and circulation. In Ampere's law, you consider the quantity $\oint \vec{B}\cdot d\vec{l}$ around the bounding path $C$. You do not consider the flux $\int \vec{B}\cdot d\vec{A}$ over the surface bounded by $C$. In your example, it looks like $\vec{B} \perp d\vec{l}$ everywhere.

1

A ferromagnet is attracted to a solenoid's magnetic field. By this I hope you mean there is already current in the solenoid. Then as you are saying magnet is attracted towards the solenoid's magnetic field, for this to happen we can have two cases. The two cases are: 1.Current in the solenoid is clockwise (viewed from the lefft or right side, say ...

1

Yes, there is an induced EMF called "Back EMF" or "Counter EMF". This is proportional to the rate of rotation of the coil; the higher the velocity of the coil the higher the counter induced EMF. The counter EMF can be calculated by subtracting the impressed voltage ($V$) in the coil from the supply EMF coming from the power source: $e = E - IR$, and \$V = ...

Only top voted, non community-wiki answers of a minimum length are eligible