Awire 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?

The magnetic induction of the magnetic field is  0.0005293 mT

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

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.