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Westkost [7]
3 years ago
14

Suppose you were asked to demonstrate electromagnetic induction. Which of the following situations will result in an electric cu

rrent?
A magnet is moving toward a wire loop.
A wire loop is moving away from a magnet.
A wire loop is rotated near a magnet.
All of the above
Physics
1 answer:
Serggg [28]3 years ago
6 0
The correct answer is "All of the above".

In fact, electromagnetic induction occurs when there is a change of the magnetic flux through the area enclosed by a circuit (in this case, the area enclosed by the wire loop).
The magnetic flux \Phi_B through a certain surface is given by
\Phi_B=B A cos \theta (1)
Where B is the intensity of the magnetic field, A is the area enclosed by the circuit and \theta is the angle between the direction of the field B and the perpendicular to the area.

In the first situation, the magnet is getting closer to the loop, so the magnetic flux through the area enclosed by the wire is increasing (because the intensity of the magnetic field B is increasing). Situation 2) is the opposite case: the wire loop is moving away from the magnet, so the intensity of the magnetic field B is decreasing, and therefore the magnetic flux is decreasing as well.
Finally, in the third situation the wire loop is rotating. Here the distance between the loop and the magnet is not changing, but remember that the magnetic flux depends also on the angle between the direction of the magnetic field and the perpendicular (formula 1), and so since the wire loop is rotating, than this angle is changing, therefore the magnetic flux is changing as well.
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A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista
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Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

C' = \frac{1}{2}C

The potential difference across the capacitor is given by

V= \frac{Q}{C}

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

V' = 2 V

Now we can analyze the electric field between the plates of the capacitor, which is given by

E=\frac{V}{d}

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

E'=\frac{2V}{2d}=\frac{V}{d}=E

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

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I hope this helps. 

Have a nice day. 
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