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Moving Charges And Magnetism

Question
CBSEENPH12037678

Using Biot-Savart’s law, derive an expression for the magnetic field at the centre of a circular coil of radius R, number of turns N, carrying current i.

Solution
As per Biot-Savart law, the magnetic field due to a current element dl at the observation point whose position vector r is given by
                dB = μ0I4π.dl × rr3
where, μ0 is the permeability of free space.

Consider a circular loop of wire of radius 'r' carrying a current 'I' and also a current element 'dl' of the loop. 



The direction of dl is along the tangent, so dl ⊥ r.

Using Biot-Savart law,

Magnetic field at the centre O due to this current element is 

              dB = μ0I4πdl sin 90°r3     = μ0I4πdlr2 

The magnetic field due to all such current elements will point into the plane of paper at the centre O.
Hence, the total magnetic field at the centre O is given by

  B = dB = μ0Idl4πr2 

  B = μ0I4πr2dl = μ0I4πr2.l 

     = μ0I4πr2.2πr  B = μ0 I2r 

For a coil of N turns, magnetic field, B = μ0NI2r .


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