Question

A piece of wire is bent to form a circle with radius r. It has a steady current I flowing through it in a counterclockwise direction as seen from the top (looking in the negative z-direction).
What is Bz(0), the z component of B at the center (i.e., x=y=z=0) of the loop?
Express your answer in terms of I, r, and constants like μ0 and π.

Answer 1

Answer:

B=μ₀I/2r

Explanation:

Produced magnetic field due to an existing electric field through a coil or conductor can be explained by Biot-Savart Law. Formula for this law is:

dB=(μ₀I/4π.r²)dL

Here,

r=Radius of the loop

I and r are constants with respect to length L.

To convert linear displacement L into angular displacement Ф:

dL=r.dФ

So,

dB=(μ₀I/4π.r²)r.dФ

dB=(μ₀I/4π.r)dФ

Integrating both sides over the circle i.e. from 0 radians to 2π radians  (360⁰), while the integration will apply only on dФ as all others are constants.

B=(μ₀I/4πr)(2π-0)

B=(μ₀I/2r)

Answer 2

The magnetic field produced due to current flowing through the coil is $B = \frac{\mu_o I}{2r}$.

The magnetic field produced due to current flowing through a coil given by Biot-Savart Law.

$dB = \frac{\mu_o I}{4\pi r^2} dL$

where;

• B is the magnetic field strength
• r is the radius of the loop
• I is the current flowing in the coil

$dB = \frac{\mu_o I}{4\pi r^2} .(rd \phi)\\\\dB = \frac{\mu_o I}{4\pi r} \ d \phi\\\\B = \frac{\mu_o I}{4\pi r} [\phi ]^{2\pi} _{0}\\\\B = \frac{\mu_o I}{4\pi r} (2\pi)\\\\B = \frac{\mu_o I}{2 r}$

Thus, the magnetic field produced due to current flowing through the coil is $B = \frac{\mu_o I}{2r}$.

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