Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/314

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282

CURRENT-SHEETS.

[678.

678.] To determine the effect of the positive end of the solenoids we must calculate the coefficient of induction on the outer solenoid due to the circular disk which forms the end of the inner solenoid. For this purpose we take the second expression for $V$ , as given n equation (15), and differentiate it with respect to $r$ . This gives the magnetic force in the direction of the radius. We then multiply this expression by $2pir^{2}\,d\mu$ , and integrate it with respect to $\mu$ from $\mu =0$ to $\mu ={\frac {z}{\sqrt {z_{2}+c_{1}^{2}}}}$ . This gives the coefficient of induction with respect to a single winding of the outer solenoid at a distance $z$ from the positive end. We then multiply this by $dz$ , and integrate with respect to $z$ from $z=l$ to $z=0$ . Finally, we multiply the result by $n_{1}n_{2}$ , and so find the effect of one of the ends in diminishing the coefficient of induction.

We thus find for the value of the coefficient of mutual induction between the two cylinders,

M=4\pi ^{2}n_{1}n_{2}c_{2}^{2}(l-2c_{1}a)

,

(21)

where	⁠	$a$	${}={\frac {1}{2}}{\frac {c_{2}+l-r}{c_{2}}}+{\frac {1\cdot 3}{2\cdot 4}}\cdot {\frac {1}{2\cdot 3}}{\frac {c_{2}^{2}}{c_{1}^{2}}}\left(1-{\frac {c_{1}^{3}}{r^{3}}}\right)$
			${}+{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\cdot {\frac {1}{4\cdot 5}}{\frac {c_{2}^{4}}{c_{1}^{4}}}\left({\frac {5}{12}}-{\frac {2}{3}}{\frac {c_{1}^{3}}{r^{3}}}+4{\frac {c_{1}^{5}}{r^{5}}}-{\frac {15}{4}}{\frac {c_{1}^{7}}{r^{7}}}\right)+\mathrm {\&c.}$ .	(22)

where $r$ is put, for brevity, for ${\sqrt {l^{2}+c_{1}^{2}}}$ .

It appears from this, that in calculating the mutual induction of two coaxal solenoids, we must use in the expression (20) instead of the true length $l$ the corrected length $l-2c_{1}a$ , in which a portion equal to $ac_{1}$ is supposed to be cut off at each end. When the solenoid is very long compared with its external radius,

a={\frac {1}{2}}+{\frac {1}{16}}{\frac {c_{2}^{2}}{c_{1}^{2}}}+{\frac {5}{768}}{\frac {c_{2}^{4}}{c_{1}^{4}}}+\mathrm {\&c.}

.

(23)

679.] When a solenoid consists of a number of layers of wire of such a diameter that there are $n$ layers in unit of length, the number of layers in the thickness $dr$ is $n\,dr$ , and we have

G=4\pi \int n^{2}dr

, ⁠

g=\pi l\int n^{2}r^{2}dr

.

(24)

If the thickness of the wire is constant, and if the induction take place between an external coil whose outer and inner radii are $x$ and $y$ respectively, and an inner coil whose outer and inner radii are $y$ and $z$ , then, neglecting the effect of the ends,

Gg={\frac {4}{3}}\pi ^{2}ln_{1}^{2}n_{2}^{2}(x-y)(y^{3}-z^{3})

.

(25)