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Question: Obtain an expression for magnetic flux density B at the center of a circular coil of radius R, havin...

Obtain an expression for magnetic flux density B at the center of a circular coil of radius R, having N turns and carrying a current II.

Explanation

Solution

This problem is based on the application of Biot-savart’s law. This law is used to determine the strength of the magnetic field at any point due to a current-carrying conductor. Biot-savart’s law for the magnetic field obeys inverse square law and the superposition principle.

Complete step by step answer:
Consider a circular coil or a circular loop of radius R, carrying current I. The circular coil consists of a large number of current elements of length dldl. Let the current element (AB) IdlId\vec l.

AS per Biot-savart’s law, the magnetic field at point C due to current element AB.
dB=μ04πIdl×rR2\Rightarrow d\vec B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{Id\vec l \times r}}{{{R^2}}}
Since the angle between dld\vec l and r^\hat r is 9090^\circ , so
dl×r^=dlsin90d\vec l \times \hat r = dl\sin 90^\circ
dl×r^=dl\Rightarrow d\vec l \times \hat r = dl
Simplifying the above equation, it gives us,
db=μo4πIdlR2db = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{{Idl}}{{{R^2}}}
Therefore, the magnetic flux at the center of a current-carrying loop
B=μoI4π×1R2dlB = \dfrac{{{\mu _o}I}}{{4\pi }} \times \dfrac{1}{{{R^2}}}\oint {dl}
B=μ0I4πR22πR\Rightarrow B = \dfrac{{{\mu _0}I}}{{4\pi {R^2}}}2\pi R
B=(μ04π)2πIR\Rightarrow B = \left( {\dfrac{{{\mu _0}}}{{4\pi }}} \right)\dfrac{{2\pi I}}{R}
However, if the direction of the magnetic field at the center of the current-carrying loop is perpendicular to the plane of the loop and in the upward direction if the current in the loop passes anticlockwise. If a coil has N turns, then the magnetic field at the current coil its center is given by
B=(μo4π)N×2πIR\Rightarrow B = \left( {\dfrac{{{\mu _o}}}{{4\pi }}} \right)\dfrac{{N \times 2\pi I}}{R}
B=μoNI2R\Rightarrow B = \dfrac{{{\mu _o}NI}}{{2R}}

Therefore, An expression for magnetic flux density B at the center of a circular coil of radius R, having N turns and carrying a current II is given by B=μoNI2RB = \dfrac{{{\mu _o}NI}}{{2R}}.

Note:
Here dbdb is the magnetic field, μo{\mu _o} is the absolute permeability and dldl is the small element length. Here the current element is a vector quantity whose magnitude equal to the product of current and length of a small element having a direction to the flow of current i.e. IdlId\vec l.