Question

Find the magnetic field of a very long solenoid, consisting of n closely wound turns per unit length on a cylinder of radius r, each carrying a steady current i

Answer 1

μ is the magnetic permeability constant; n is the turn density N/L; I is the current running through the solenoid. Note that the magnetic field outside of the solenoid is 0 . ... Consider a very long solenoid, consisting of n closely wound turns per unit length on a cylinder of radius R , each carrying a steady ...

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Answer 2

Ampère’s law gives:

$\oint{\pmb{B}}.d\pmb{I}=[B(a)-B(b)]L=\mu _{0}I_{enc}$

since loop encloses no current,

• So, magnetic field of the infinite, closely wound solenoid run parallel to  axis. From right-hand rule, we can expect that it points upward inside  solenoid & downward outside.

• Moreover, it certainly approaches zero as we go very far away. With this , let us apply Ampere’s law to the 2 rectangular loops. Loop 1 lies entirely outside of the solenoid, with its sides at distances a & b from the axis:

$\oint{\pmb{B}}.d\pmb{I}=[B(a)-B(b)]L=\mu _{0}I_{enc}$=0

B(a) = B(b)

• Evidently  field outside does not depend on the distance from  axis. But we agreed that it goes to zero to large s. It must therefore be 0 everywhere.

As for loop 2, which is half inside & half outside, Ampère’s law gives

$\oint{\pmb{B}}.d\pmb{I}=BL=\mu _{0}I_{enc}=\mu _{0}nIL,$

• where B is the field inside that solenoid. (The right side of  loop contribute nothing, since B = 0 out there.) Conclusion:

B  =              $\mu _{0}nIZ,$           inside     the     solenoid,                        

                     0               outside    the    solenoid.

Notice that  field inside is the uniform,

• it does not depend on distance from  axis. In this case the solenoid is to magnetostatics what the parallel-plate capacitor is  electrostatics:

• a simple device for producing strong uniform fields.

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