Question

# Use Biot-Savart law to derive the expression for the magnetic field on the axis of a current carrying circular loop of radius R. Draw the magnetic field lines due to a circular wire carrying current I.

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Solution

## Imagine a circular coil of radius R with centre O.Let the current flowing through the circular loop be I. suppose P is any point on the axis at a distance of r from the centre 0. Let the circular coil be made up of a large number of small elements of current, each having a length of dl. According to Biot-Savart's law, the magnetic field at Point P will be dB=μ0I4π×|dl×r|r3 where, r2=x2+R2 |dl×r|=rdl [ ∵ Both are perpendicular] Here, r is the position vector of point O from the current element. dB has two components i.e.,dBx and dBy. dBy is cancelled out and only the x-component remains. ∴dBx=dB cosθ cosθ=R√x2+R2 dBx=μ0Idl4π.R(x2+R2)32 But, ∫dl=2πR So, B=μ0IR×2πR4π(x2+R2)32 For n turns in the circular loop, →B=μ0nIR22(x2+R2)32.^i

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