Field Strength Due To A Solenoid

Let us consider a solenoid of radius a metres and length l metres having N turns.

Consider an element of very small length dx, then turns on this small element will be  Ndx .
                                                                                                                           l

Therefore field strength at the centre of the solenoid due to small element dx, which may be considered as a coil having  Ndx turns, is given by 
                         l
dH = 1a² X Ndx = NI . a²  dx
         2r³     l       2l     r³

and From cot θ = x/a
Differentiating both sides we get

-cosec² θ dθ = dx
                       a

or  dx = -a cosec² θ dθ = -adθ
                                    sin²θ

a² dx = a² x 1x dx = sin² θ x 1 x -adθ = -sin θ dθ        sin θ = a
r²         r²     a                      a   sin²θ                                   r

Substituting a²  dx = - sin θ dθ in expression (i) we get

dH = NI . (sin θ ) dθ = -NI sin θ dθ
       2l                       2l

Integrating both sides we get



If the solenoid is very long, ∝ is approximately equal to zero and ß is approximately equal to π and, therefore, ∝ = 1

Hence field intensity, H = NI [1-(-1)] = NI  AT/m.
                                    2l                l

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