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480. Problem 35.56 (RHK) We have to derive the solenoid equation starting from the expression for the field on the axis of a circular loop. |
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Solution: Click For PDF Version Let the number of turns per unit length of the solenoid be
We recall that the magnetic field at the axis of a circular current carrying loop at a distance x from its centre is given by
Its direction is along the axis perpendicular to the plane of the circular loop. We will calculate the magnetic field at P, which we take to be the origin of
the x-axis. Let us consider a section of the solenoid of thickness
Therefore, the magnetic field at P at the axis of the solenoid will be given
by integrating
This integral is readily evaluated by making the substitution
We find
The magnetic field at the axis of the solenoid is, therefore,
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We will calculate the magnetic field at the axis of an ideal solenoid. That is
we assume that the solenoid is of infinite length. Let the radius of the
solenoid be r. 
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at a distance x from the point P. Let the current flowing in the
circular loops of the solenoid be 
with
respect to x from
to
.
We have
