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2 years ago

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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?a. B1, b. B1 / 4, c. 4B1, d. B1 / 2

1 answer:

soldi70 [24.7K]2 years ago

4 0

Answer:

D.

Explanation:

To solve the exercise it is necessary to apply the concepts related to the Magnetic Field described by Faraday.

The magnetic field is given by the equation:

$B = \frac{\mu_0 I}{2\pi d}$

Where,

$\mu =$ Permeability constant

d = diameter

I = Current

For the given problem we have a change in the diameter, twice that of the initial experiment, therefore we define that:

$B_1 = \frac{\mu_0 I}{2\pi d}$

$B_2 = \frac{\mu_0 I}{2\pi 2d}$

The ratio of change between the two is given by:

$\frac{B_2}{B_1} = \frac{\frac{\mu_0 I}{2\pi d}}{\frac{\mu_0 I}{2\pi 2d}}$

$\frac{B_2}{B_1} = \frac{d}{2d}$

$\frac{B_2}{B_1} = \frac{1}{2}$

$B_2 = B_1 \frac{1}{2}$

Therefore the correct answer is D.

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