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levacccp [35]
3 years ago
13

An experiment is conducted on a long straight wire of diameter d. A constant current is sent through the wire and the magnetic f

ield on the surface of the wire is measured to be B1. The experiment is then repeated with the same current but with a wire of diameter 2d. The magnetic field measured on the surface of this second wire will be which of the following?
B1, B1 / 4, 4B1, B1 / 2
Physics
1 answer:
Anton [14]3 years ago
5 0

Answer:

B_2 = \frac{B_1}{2}

Explanation:

As per Ampere's law the magnetic field at the surface of the wire is given as

\int B. dL = \mu_0 i

here we have

B . (\pi d) = \mu_0 i

so we will have

B_1 = \frac{\mu_0 i}{\pi d}

now again we use same value of current but wire with double the diameter

so the magnetic field at the surface is given as

B_2 = \frac{\mu_0 i}{2\pi d}

so we have

B_2 = \frac{B_1}{2}

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Hello, I can help you with this

The force that the earth exerts on any object is due to the force of gravity and is known as weight, F = mg, this force is directed towards the center of the earth, in addition, according to the laws of physics for any action must have a reaction, in this case it is a force that goes out  from the ground and it is directed upwards, this force is called normal and it has the same magnitude of your weight but the opposite direction.

Have a great day.

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