Magnetic Field at the centre of Spiral

[AGNuke]

A thin insulated wire forms a plane spiral of N = 100 tight turns, carrying a current I = 8 mA. The radii of inside and outside turns are equal to a = 50 mm and b = 100 mm.

Find the magnetic induction at the centre of the spiral and the magnetic moment of the spiral with the given current.

This problem I searched all over but nowhere I could found any answer. Hope I'll be helped.

The figure of planar spiral is given with inner radius and outer radius as mentioned, in case anybody needs a diagram.

I first calculated the no. of turns per unit width, that would be $$n=\frac{N}{(100-50)mm} = 2000\;turns/m$$

I know the magnetic field due to a current carrying loop with n turns,$$B = \frac{\mu _0ni}{2r}$$

I considered a loop out of spiral of radius r and width dr, thus the total magnetic field would be the summation of all the loops from inner radius to outer radius.$$\int \mathrm{d}B=\int_{0.05}^{0.1} \frac{\mu _0ni}{2r}\mathrm{d}r$$

I tried to calculate but I dodged the answer from a long mile, the answer is $7\; \mu T$

Any Help appreciated. And if anybody do not mind, I would also like to get my question asked no too long ago answered too. https://www.physicsforums.com/showthread.php?t=641032

 

[TSny]

I first calculated the no. of turns per unit width, that would be $$n=\frac{N}{(100-50)mm} = 0.002\;turns/m$$

Watch your unit conversion here. Otherwise, everything looks good.

 

[AGNuke]

Its not about unit conversion. I had all done correct. The expression after integration was including ln2, which is not present in the answer.

Even if I substituted the value of pi or ln2, my answer was really off.

 

[TSny]

Your answer for n says that there are only .002 turns of wire in a radial distance of 1 meter. But you know that you have 100 turns in only 50 mm of distance.

 

[AGNuke]

Extremely sorry. Its 2000 turns/m. And the answer I got is 6.4 microTesla.$$B=\frac{\mu _0ni}{2}\mathrm{ln}2$$

 

[TSny]

Extremely sorry. Its 2000 turns/m. And the answer I got is 6.4 microTesla.$$B=\frac{\mu _0ni}{2}\mathrm{ln}2$$

Your expression for the answer looks correct. I don't get 6.4 microTesla.

 

[AGNuke]

Yeah... I got something like 6.96. Maybe my calculator was malfunctioning and I didn't bothered to recalculate it on Computer. Fault on my part.

Can you look at the question I posted long before, unfortunately no one has entertained it as of yet - https://www.physicsforums.com/showthread.php?t=641032