Question

Derive the expression for magnetic field intensity of a long solenoid?

Answer 1

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Self Inductance of a Current Carrying Solenoid: Consider a solenoid of radius r metre, length l metre, having total number of turns N and carrying a current I ampere. The magnetic field produced on its axis inside the solenoid is

B=lμ0NINA−1m−1

If it is assumed that the solenoid is very long, the intensity of magnetic field B can be assumed to be uniform inside it at each point, then the total magnetic flux linked with the solenoid is

B=4πμ0m[{(d−l)(d+l)}2d2+l2+2dl−(d2+l2−2dl)]

ϕ=B× total effective area of solenoid

   =B×(NA)

where A=πr2 is the area of cross section of solenoid.

∴ϕ=lμ0NI×NA=lμ0N2AI weber

But by definition ϕ=LI where L is the self-inductance of solenoid, therefore

Self inductance L=Iϕ=lμ0N2A henry

If inside the solenoid instead of air or vacuum there is a substance of relative permeability

L=lμ0μrN2A henry

Thus, the self inductance of a solenoid depends on the following factors:

(i) On the number of turns in the solenoid: The self inductance increases on increasing the number of turns.

(ii) On the area of cross section of solenoid (i.e., on the radius of solenoid): The self inductance increases on increasing the area of cross section.

(iii) On the length of solenoid: On increasing the length of solenoid, its self inductance decreases.

(iv) On the relative permeability of the substance placed inside the solenoid: If a soft iron rod is placed inside the solenoid, its self inductance increases.

Remember that if the number of turns per unit length in the solenoid is n, then n=N/l and the self inductance of solenoid is

L=μ0n2lA henry

Answer 2

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