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A wire with radius 23 cm has a current of 7 A which is distributed uniformly through its cross sectional area. If you were to use ampere’s law to calculate the magnetic field at a distance of 2 cm from the center of the wire, what would be the current enclosed?

User Yewge
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1 Answer

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Answer:

The magnetic induction of the magnetic field is 0.0005293 mT

Step-by-step explanation:

Data given

I = 7 A = the total current in the wire

r = 23 cm = the radius of the wire = 0.23 meter

r' = 2cm = the measurement point, which should be inside the wire = 0.02 meter

Let's consider the current density is constant in the wire, ⇒ the current enclosed is a function of the enclosed area

I(enclosed) = Jπ r ²

we can consider the current density as the total current over the whole area:

I(enclosed) = I / (πr ²) * πr' ²

I(enclosed) = (I* r'²)/ (r ²)

with I = total current in the wire = 7A

With r = the radius of wire = 0.23 meter

with r' = the distance of point from the center of wire 0.02 meter

We plug this into ampere's law:

∮ *B *dl =μ 0 * (I* r'²)/ (r ²)

with B = Magnetic flux density (in Tesla) or magnetic induction

with dl = an infinitesimal element (a differential) of the curve C

with µ0 = the magnectic constant = 4π*10^−7 H/m

We can simplify this, by using an Amperian loop can write this as:

B *( 2 π r') = μ 0 * (I* r'²)/ (r ²)

Because the circumference of a circle is 2 π r , when we integrate over length at a distance r ′ from the center of wire whose crossection is a circle we get 2 π r ′

When we isolate B, we get:

B = µo *(Ir'/2 π r ²)

B = 4π*10^−7 * ((7*0.02)/2*π*0.23²)

B =5.293 *10 ^-7 T = 0.0005293 mT

The magnetic induction of the magnetic field is 0.0005293 mT

User Jan Baer
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