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4 years ago

A student creates an electromagnet, How could the student create a larger current in the electromagnet?

Physics

1 answer:

antoniya [11.8K]4 years ago

4 0

Answer

When an electric current flows in a wire, it creates a magnetic field around the wire. This effect can be used to make an electromagnet. A simple electromagnet comprises a length of wire turned into a coil and connected to a battery or power supply.

A coil of insulated wire is wrapped around an iron nail. One end of the coil of wire is connected to a battery. The other is connected to a switch in the circuit

A simple electromagnet

You can make an electromagnet stronger by doing these things:

wrapping the coil around a piece of iron (such as an iron nail)

adding more turns to the coil

increasing the current flowing through the coil

There is a limit to how much current can be passed safely through the wire because the resistance of the wire causes heating.

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Suppose you had an electrical circuit and decided to change it into a new circuit. does the equivalent resistance of the new cir

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Answer:

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Explanation:

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2 years ago

Heat is an exchange of internal energy from one system to another due to

aleksandrvk [35]

a temperature difference between the systems.

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3 years ago

Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying

agasfer [191]

Answer:

The charge density in the system is $4.25*10^4C/m$

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

$r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C$

We need to asume here the number of free electrons in a copper conductor, at which is generally of $8.5 *10^{28}m^{-3}$

The equation to find the current is

$I = VenA$

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

$I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)$

$I= 22.11a$

Now to find the linear charge density, we know that

$I = \frac{Q}{t} \rightarrow Q = It$

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

$V_d = \frac{l}{t} \rightarrow l = V_d t$

The charge density is defined as

$\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}$

Replacing our values

$\lambda = \frac{22.11}{5.20*10{-4}}$

$\lambda= 4.25*10^4C/m$

Therefore the charge density in the system is $4.25*10^4C/m$

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4 years ago

A student found the rock shown above and weighed it to determine its mass. What steps should the student take to find its densit

musickatia [10]

find its density is the key so good luck buddy

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3 years ago

Waves diffract the most when their wavelength is?

Margarita [4]

<h2>Answer: about the same size of the gap </h2>

Diffraction happens when a wave (mechanical or electromagnetic wave, in fact, any wave) meets an obstacle or a slit .When this occurs, the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.

Note that the principal condition for the occurrence of this phenomena is that <u>the obstacle must be comparable in size (similar size) to the size of the wavelength. </u>

<u></u>

In other words, <u>when the gap (or slit) size is larger than the wavelength</u>, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs and the waves spread out greatly.

Therefore:

<h2>Waves diffract the most when their wavelength is <u>about the same size of the gap</u> </h2>

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4 years ago