Set the ratio of the current I' enclosed by circle 2 to the entire current I equal to the ratio of the area πr2 enclosed by circle 2 to the cross-sectional area πR2 of the wire:

Question

11-03-2020
Physics

contestada

Set the ratio of the current I' enclosed by circle 2 to the entire current I equal to the ratio of the area πr2 enclosed by circle 2 to the cross-sectional area πR2 of the wire:

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lcoley8 lcoley8 · Answer 1 · 2020-03-16T06:06:19+01:00

Answer:

This exercise is incomplete, therefore the complete exercise says so: A long, straight wire of radius R carries a steady current I that is uniformly distributed through the cross section of the wire. Calculate the magnetic field a distance r from the centre of the wire in the regions r>=R and r<R

The answer are

B = (μo*I)/2pi*r, this is for r>= R

B = ((μo*I*r) / (2pi*R^2)), this for r<R

Explanation:

The total current that runs through the circle is equal to I and if we use Ampere's law we are left with an expression:

∮B ds = B∮ds = B(2pi*r) = μo*I

Solving for B, we have:

B = (μo*I)/2pi*r, this is for r>= R

If the interior of the cable, r<R, the current I´ that runs through circle 2 is less than the total current I. The equation of the current I´ that is inside circle 2 to the total current I is equal to the area pi*r^2 within circle 2 and the cross-sectional area pi*r^2 of the cable, as follows:

I´/I = (pi*r^2)/pi*R^2

Clearing I´:

I´ = (r^2*I)/R^2

Applying Ampere's law:

∮B ds = B (2pi*r) = μo*I´ = μo * ((r^2*I)/R^2)

Solving for B, we have the requested expression:

B = ((μo*I*r) / (2pi*R^2)), this is for r<R