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3 years ago

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

Physics

1 answer:

antoniya [11.8K]3 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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3 years ago

What is the total surface charge qint on the interior surface of the conductor (i.e., on the wall of the cavity)

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Answer: hello your question is incomplete below is the missing part

A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.

answer:

- q

Explanation:

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3 years ago

How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800

ozzi

Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

$A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2$

The inductance of the solenoid is given by;

$L = \frac{\mu_0 N^2A}{l} \\\\L = \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H$

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

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3 years ago

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svetoff [14.1K]

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And I hope this helped :)

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3 years ago