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  • Derive an expression for the axial magnetic field of a finite solenoid of length 2l and radius r carrying current I. Under what condition does the field become equivalent to that produced by a bar magnet?   <div style="float:left;

Derive an expression for the axial magnetic field of a finite solenoid of length 2l and radius r carrying current I. Under what condition does the field become equivalent to that produced by a bar magnet?

 

 
 
 
 
 

Answers (1)

The magnetic field at P due to small element dx is 

dB = \frac{\mu _{0}ndxIa^{2}}{2\left [ \left ( r-x \right )^{2}+a^{2} \right ]^{\frac{3}{2}}}

for r>>>a and r>>> x

B\approx \frac{\mu _{0}nIa^{2}}{2r^{3}}\int_{-l}^{l}dx

dB= \frac{\mu _{0}nIa^{2}2l}{2r^{3}}

Magnetic moment

 m = n2l(\pi a^{2})

\therefore B = \frac{\mu_{0}2m}{4\pi r^{3}}

The value of B obtained is similar to that of a bar magnet at an axial point far away from it.

Posted by

Safeer PP

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