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Derive an expression of the magnetic field at the center of a circular current carrying coil.

Answer
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Hint: A wire carrying current has a magnetic field. The intensity of the magnetic field at any point is obtained by the Biot-Savart’s law.
This law in vector form can be written as
dB=μ04πidl×r^r2

Step by step solution:
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1) Consider a current carrying circular loop having its center at O carrying current i. If dl is a small element at a distance r then, the magnetic field intensity on that point can be written using Biot-Savart’s law.
dB=μ04πidl×r^r2|dB|=μ04πidlsinθr2
2) As the loop is circular then,
θ=90
sinθ=1
3) Putting this in the Biot-Savart’s law we will get,
|dB|=μ04πidlr2
4) If we think that the circular loop is composed of numbers of such small element dl, then we will get the magnetic intensity for all over the loop. So to get the total field we must sum up that is integrate the magnetic field all over the field.
B=0BdBB=μ04πidlr2B=μ04πir2dl
5) As the integration over dl returns the circumference of the loop then, you can write,
dl=2πrB=μ04πir22πrB=μ0i2r
Hence the magnetic field at the center of the circular coil is B=μ0i2r.

Note: Field due to a semi-circular coil is just half that of the circular coil.
B=μ0i4r