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  • 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 due to a circular wire carrying current I.     <div style=

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 due to a circular wire carrying current I.
 

 

 

 

 
 
 
 
 

Answers (1)


Consider two small current elements on the circular loop which are opposite to each other.
then the magnetic field at a point p which is at a distance X from the centre of the coil is due to the x component of the fields of A and B and y components cancel out.
\because dB_{X}= \frac{\mu_{0}}{4\pi }\frac{Idl\times \sin \left ( 90 \right )}{\left ( \sqrt{R^{2}+X^{2}} \right )^{2}}\cos \theta
  dB_{X}= \frac{\mu_{0}Idl}{4\pi }\times \frac{1}{\left ( \sqrt{R^{2}+X}^{2} \right )^{2}}\times \frac{R}{\sqrt{R^{2}+X^{2}}}
            = \frac{\mu_{0}Idl}{4\pi }\frac{R}{\left ( X^{2}+R^{2} \right )^{\frac{3}{2}}}
B= \int dBX= \frac{\mu_{0}IR}{4\pi \left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}\int dl
   = \frac{\mu_{0}I\times 2\pi R^{2}}{4\pi \left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}
   = \frac{\mu_{0}IR^{2}}{2\left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}

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Safeer PP

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