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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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Answer:

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b) <em>1.139 x 10^-6 Ampere</em>

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Explanation:

The radius of the wire r = 2 mm = 0.002 m

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direction of the magnetic field ∅ = 25°

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varying time = 2 sec

The cross sectional area of the wire = \pi r^{2}

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The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>

<em></em>

<em></em>

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

I = E/R

where R is the resistor

I = (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>

<em></em>

<em></em>

<em> </em>c) power delivered to the resistor is given as

P = IE

P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>

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