A short solenoid (length l and radius a, with n1 turns per unit length) lies on the axis of a very long solenoid (radius b, n2 turns per unit length) as shown in Fig. 7.32. Current I flows in the short solenoid. What is the flux through the long solenoid?

Question

A short solenoid (length l and radius a, with [latex]n_1[/latex] turns per unit length) lies on the axis of a very long solenoid (radius b, [latex]n_2[/latex] turns per unit length) as shown in Fig. 7.32. Current I flows in the short solenoid. What is the flux through the long solenoid?

Accepted Answer

Since the inner solenoid is short, it has a very complicated field; moreover, it puts a different flux through each turn of the outer solenoid. It would be a miserable task to compute the total flux this way. However, if we exploit the equality of the mutual inductances, the problem becomes very easy. Just look at the reverse situation: run the current I through the outer solenoid, and calculate the flux through the inner one. The field inside the long solenoid is constant:

[latex]B = μ_{0}n_{2} I[/latex]

(Eq. 5.59), so the flux through a single loop of the short solenoid is

[latex]B= \begin{cases} μ_{0}nI \hat{z}, inside the solenoid, \μ_{0}nI ˆz, inside the solenoid,\end{cases} [/latex] (5.59)

[latex]B\pi a^{2} = μ_{0}n_{2} I\pi a^{2}.[/latex]

There are [latex]n_{1}l [/latex] turns in all, so the total flux through the inner solenoid is

[latex]\phi = μ_{0}\pi a^{2}n_{1}n_{2}l I.[/latex]

This is also the flux a current I in the short solenoid would put through the long one, which is what we set out to find. Incidentally, the mutual inductance, in this case, is

[latex]M = μ_{0}\pi a^{2}n_{1}n_{2}l .[/latex]

Question 7.10: A short solenoid (length l and radius a, with n1 turns per u...

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