Magnetic field created by a solenoid

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SUMMARY

The discussion centers on calculating the magnetic field generated by a solenoid surrounding a long, straight, hollow plastic tube containing ionized liquid. The magnetic field on the axis of the tube is determined to be 0 T, as confirmed by multiple participants. However, an alternative approach considers the system as a dipole, leading to the formula B = (μ₀ / 2πd³)Nπr², where μ₀ is the permeability of free space, N is the number of loops per meter, and r is the radius of the tube. Participants debate the validity of this dipole approximation and the inclusion of the ionic current in the calculations.

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Pepealej
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Homework Statement

Inside a very long, straight and hollow plastic tube (R << L) we have ionized liquid moving at a velocity v = 100 m/s. The net charge per unit volume is n = 106 electrons/m3. The radius of the tube is R = 10cm. Around the tube we have a solenoid, with n = 100 loops/m, through which circulates a current I = 2 A.

Estimate the magnetic field on the axis of the tube at a distance d >> L.

Book's answer

0 T or, considering the system as a dipole, [tex]B=\dfrac{\mu_0}{2\pi d^3}N\pi r^2[/tex]

My question

I understand how the magnetic field is 0 T but, how can I consider the system as a dipole to obtain the other answer?

Thanks in advance!
 
Last edited:
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Pepealej said:
Book's answer

0 T or, considering the system as a dipole, [tex]B=\dfrac{\mu_0}{2\pi d^3}N\pi r^2[/tex]

My question

I understand how the magnetic field is 0 T but, how can I consider the system as a dipole to obtain the other answer?

Thanks in advance!


Well, we can all see B = 0. But their other answer (dipole) makes no sense since neither the ion nor the wire current I is included in their formula. Probably a typo, but ...

I would say:
Consider all the turns of the coil bunched together to form a single-turn loop with current = nLI. That makes using Biot-Savart easy along the axis & is justifiable by d >> L.

As for the ionic current, about all I could see is trying to compute the magnetic vector potential integral A, then B = del x A.
 

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