Question

The general solution of
d2B/dr2+dB/rdr-B/a^2 =0
is B=C Io(r a) +D Ko(r/a)
In our particular case the solution is Ko.
How we get
B= (h/2 phi a^2) Ko(r/a)
Where h= magnetic flux= Integral B d2r. Phi=3.14. And a=(m/meu. n e^2) 1/2

Answer 1

Answer on Question #63012 – Math – Differential Equations

Question

The general solution of

$$\mathrm{d}2\mathrm{B}/\mathrm{d}r2 + \mathrm{dB}/\mathrm{r dr}-\mathrm{B}/a^2 = 0$$

is $B = C \operatorname{Io}(r a) + D \operatorname{Ko}(r/a)$

In our particular case the solution is Ko.

How we get

$$B = (h/2 \phi a^2) \operatorname{Ko}(r/a)$$

Where $h =$ magnetic flux $=$ Integral B d2r. Phi $= 3.14$. And $a = (m/ \text{meu. } n \, e^2) \frac{1}{2}$

Solution

If the solution has to be $B = D K_0 \left( \frac{r}{a} \right)$, then constant of integration usually can be found by applying boundary conditions. But if boundary conditions are not given, then we can find D as the mean value of B passing through the given region. If the region is a circle with the radius $a$, then the mean value is

$$B_m = \frac{\int B \, dS}{S} = \frac{\int B \, dS}{\pi a^2}$$

Then we get:

$$B = \frac{\int B \, dS}{\pi a^2} K_0 \left( \frac{r}{a} \right)$$

Answer: $B = \frac{\int B \, dS}{\pi a^2} K_0 \left( \frac{r}{a} \right)$

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