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Thepotemich [5.8K]
2 years ago
13

What the name for the phenomenon of inducing voltage by changing the magnetic field around a conductor?

Physics
1 answer:
8090 [49]2 years ago
8 0

The phenomenon of inducing voltage by changing the magnetic field around a conductor will be Electromagnetic Induction. Option B is correct.

<h3>What is the Faraday law of electromagnetic induction?</h3>

According to Faraday's law of electromagnetic induction, the rate of change of magnetic flux link with the coil is responsible for generating emf in the coil to result in the flow of amount of current .

So in order to increase the current, we need to increase the EMF;

so we can increase it by;

1) Increasing the number of turns

2) Increase the area of the loop

3) By moving the magnet faster

Hence, option B is correct.

To learn more about Faraday law of electromagnetic induction:

brainly.com/question/13369951

#SPJ1

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What happens when a magenta light is shone on a green surface?
Sedbober [7]

Answer:

When Magenta light is shown on a green surface, it looks black.

Explanation:

It absorbs the Magenta light and also reflects none of the light.

5 0
3 years ago
Animals need oxygen to complete the many chemical reactions taking place in the body. An animal inhales oxygen and exhales _____
mezya [45]
Carbon dioxide and nitrogen
5 0
3 years ago
Read 2 more answers
A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir
mario62 [17]

Answer:

a . 0.35cm

b.  11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

8 0
3 years ago
The moon Phobos orbits Mars
podryga [215]

Explanation:

For a circular orbit v= \sqrt{\frac{G.m}{r} } with G = 6.6742 × 10^{-11}

Given m = 6.42 x 10^23 kg and  r=9.38 x 10^6 m

=> v = 2137.3 m/s

I hope this is the correct way to solve

3 0
3 years ago
A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica
Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed \omega =40rad/s

mass of m_2=0.8 kg dropped at r=0.3 m from center

let \omega _2 be the final angular velocity of cylinder

Conserving Angular momentum

L_1=L_2

\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2

\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2

26.01\times 40=26.082\times \omega _2

\omega _2=39.88 rad/s

3 0
3 years ago
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