A piece of wire is bent to develop a circle via radius r.It has a stable current I flowing via it in a counterclockwise direction as watched from theheight (looking in the negative z direction).

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What is B_z(0), the z component of vecB at the center (i.e., x=y=z=0 ) of the loop? Express your answer in regards to ^I, r,and constants prefer mu_0 and pi.

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Answer

Consider a current facet dl on the wire. The magnetic area at the facility of the loop due to the present aspect mathrmdl is,

d B=fracmu_0 I d vecl imes hatr4 pi r^2

Here, r is the radius of the loop, I is the current with the wire, and also hatr is the unit vector along the radius r.

Substitute d l sin heta for d vecl imes hatr in equation d B=fracmu_0 I d vecl imes hatr4 pi r^2.

d B=fracmu_0 I d l sin heta4 pi r^2

Substitute 90^circ for heta in equation d B=fracmu_0 I d l sin heta4 pi r^2.

d B=fracmu_0 I d l sin 90^circ4 pi r^2

=fracmu_0 I d l4 pi r^2

Explanation

The vector product of two vector quantities is, vecA imes vecB=A B sin heta

As the magnitude of unit vector is 1 , the vector product of existing element dl and hatr is,

d vecl imes hatr=d l sin heta

The direction of present element is very same as the direction of existing. The present facet and unit vector along the radius of the loop are perpendicular to each other for all the points on the loop. Hence, the angle in between the current element and also unit vector along the radius of the loop is constantly 90^circleft( heta=90^circ ight).

The direction of current is counterclockwise and then by best hand also dominance the direction of magnetic field is along the z direction. Thus, the z component of magnetic field is exact same as the net magnetic field.

Determine the magnetic field at the facility of the existing loop.

The expression for magnetic field is,

d B=fracmu_0 I d l4 pi r^2

Integprice the expression d B=fracmu_0 I d l4 pi r^2.

int d B=int fracmu_0 I d l4 pi r^2

B=fracmu_0 I4 pi r^2 int d l

Substitute 2 pi r for oint d l in equation B=fracmu_0 I4 pi r^2 oint d l

B=fracmu_0 I4 pi r^2 2 pi r

=fracmu_0 I2 r

The z component of magnetic field at the center of the loop is fracmu_0 I2 r.

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Explanation

As the existing flowing via the wire is constant, the existing mathrmI deserve to be taken out of the integral. The integral over the size aspect over a circular path gives the circumference of the circle and also the circumference of a circle of radius r is 2 pi r. Hence, the integral of length element over the circular loop is,

int d l=2 pi r

Here, r is the radius of the circle and mathrmd mathrml is the readjust in size.