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If both bulbs have identical normal tungsten filaments which are rated to run directly from the given battery, bulb B will come on bright but quickly dim because the resistance of tungsten increases with temperature. The inductor limits the rate at which the current increases in bulb A. Its brightness will start low and come up fairly fast. Since the ...
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I am proving that the area-sweeping technique gives the same result as the Lorentz force method. Using a battery across the two rods in parallel doesn't change the idea as we shall see.
Magnetic Flux $\phi=\int_A \mathbf{B}.d\mathbf{A}$
Faraday's law of electromagnetic induction transforms as follows:
\begin{align*}
\text{EMF }\varepsilon&=-\frac{d\phi}{...
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The key here is the "ground". According to your picture, you can see immediately that there is an electric field from the two rods acting (in the downward direction) on the wire connecting between the sphere and the ground. This electric field will 'kick' electrons in the wire, and generate current. The sphere will get charged after a while (it ...
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However I'm not able to explain why the electrical power has to be
constant during stepping up and down of voltage.
For an ideal (lossless) transformer energy must be conserved, i.e., power into the primary = power out of the secondary, or P=VI= constant. A transformer doesn't generate electrical energy.
Or put in other words , How can you prove Current is
...
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What do you mean by "during stepping up and down"? "Stepping up/down" isn't a transient state in the operation of a transformer. A step-up transformer continuously produces a voltage that is higher than the input voltage: it is always stepping up voltage as long as the transformer is in operation.
If the input and output powers were ...
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Yes, the charge will move under the influence of the induced E-field, but it will not move in a circle: this would require a centripetal force, but there is none. Instead, the (positive) charge will start to move in the direction of the E-field, then spiral outward.
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We know that the inductor doesn't allow the instantaneous change of
current and flux through it, so the currents in L1 and L2 should
remain the same immediately after the switch is opened up, but that is
also not possible as the direction of current in the loop containing
L1 and L2 cannot be in the same direction, so we reach to a
contradiction. How do we ...
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Your intuition is correct for these reasons:
an air conditioner typically pumps heat from one stream of air, into another stream of air. Sometimes the stream being heated goes from outside to outside, while the stream being cooled goes from inside to inside. Alternatively, the stream being heated can go from inside to outside, while the stream being ...
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You text is not very readable. In the rest frame of the magnet the magnetic field is not changing and there is no electric field. In other words no work is done on the particle.
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A fundamental reason that a static magnetic field cannot change the speed of a charge is that the force exerted by the field is always perpendicular to the velocity of the charge.
Your question seems to be different from what the title indicates: it seems that actually you are want to know if the electric field induced by a time-varying magnetic field can ...
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First, here's an example of a circuit that works as you describe.
We say that the switch has been open a long time which is to say that, before the switch closes at $t=0$, the circuit is in DC steady state (all voltages and currents are constant in time).
Since the current through the inductor is constant, the magnetic flux threading the inductor is ...
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The inductor cannot win!
When the current changes so does the magnetic flux linked with the inductor, an emf is induced which produces a current in opposition to the changing current producing it - Faraday and Lenz.
You get an endless sequence if the inductor stopped the current changing, no emf would be induced and there would be no opposition to any ...
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Objects with mass tend to resist (due to inertia) changes in their velocity. But that doesn't mean that nothing ever experiences acceleration. It's the same story here.
When we say "an inductor doesn't like changes in current" or "an inductor opposes changes in current" we don't mean that the current through the inductor can't change at ...
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Yes, it will.
You are thinking that with balanced electrostatic forces, there is no force to make electrons move. That's reasonable.
But when electrons can move from an area of high potential to an area of low potential, some of them will randomly do that. And they are less likely to move back to the high potential area than others are to move away.
When ...
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"I think that magnetic flux through conductor remains constant as B is constant."
It's not the flux "through the conductor" that matters. It's the flux through the area swept out by the conductor. Imagine that the straight conductor (length $\ell$) is lying on a table, and that there is a uniform magnetic field acting downwards. (Actually ...
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I've actually never seen this form of the law before, so if you have a reference, please let me know. I just worked out how to derive it from the "standard" form of Faraday's Law you've given at the start of your question, but I'm going to make a couple of assumptions, let me know if they aren't justified:
The magnetic field is constant in space, ...
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The Gauss's Law applies to a CLOSED Surface. What does it means?
Imagine a sphere. The sphere has a inside and a outside. If something is in the inside, it must cross the surface in order to get outside.
It means that all closed surfaces divides the space in two regions and you need to cross the surface to go from one region to the other.
What Gauss's Law ...
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