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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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Answer

given,

given,

small cube side = 10 cm

larger cube side = 12 cm

density of steel = 7 g/cm³

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buoyant force of Cube A

 B₁ = ρ₁ V g = 1 x 10 x 10 x 10 x g=  1000 g

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buoyant force acting on the cube depends on the density of the fluid

hence,

B₂ = B₄  > B₁ = B₃

 

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Answer:

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Explanation:

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