Question
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 I.
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 dl. Let the current element (AB) Idl.
AS per Biot-savart’s law, the magnetic field at point C due to current element AB.
⇒dB=4πμ0R2Idl×r
Since the angle between dl and r^ is 90∘, so
dl×r^=dlsin90∘
⇒dl×r^=dl
Simplifying the above equation, it gives us,
db=4πμoR2Idl
Therefore, the magnetic flux at the center of a current-carrying loop
B=4πμoI×R21∮dl
⇒B=4πR2μ0I2πR
⇒B=(4πμ0)R2πI
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=(4πμo)RN×2πI
⇒B=2RμoNI
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 I is given by B=2RμoNI.
Note:
Here db is the magnetic field, μo is the absolute permeability and dl 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. Idl.