Magnetic Field Due To Wire Of Semi-Circular Cross Section

[Aurelius120]

Homework Statement

An infinite wire has semi-circular cross section with radius R. It carries current I. Then the magnitude of magnetic induction along its axis

Relevant Equations

Field due to infinite wire=$\frac{\mu_○I}{2\pi d}$

I proceeded as follows
Current in sector $d\theta=$ is:
$$dI=\int_{x=0}^{x=R}{\frac{I}{\pi R^2/2}\times\frac{d\theta}{2\pi}\times 2\pi x dx}$$
Field due to sector $d\theta$ is therefore
$$dB=\int^{x=R}_{x=0}{\frac{\mu_○}{2\pi x}\times\frac{I}{\pi R^2/2}\times\frac{d\theta}{2\pi}\times 2\pi x dx}$$
$$\implies dB=\frac{\mu_○IR}{\pi^2R^2}d\theta$$

For the entire wire:
$$B_{net}=\int^{\pi}_0{dB \cos(\theta-\pi/2) \hat i} + \int^{\pi}_0{dB \sin(\theta-\pi/2) \hat j}$$
Finally$$ B_{net}=\frac{2\mu_○I}{\pi^2R}$$

This is double the correct value? What did I do wrong ? Why does the book give this answer?

 

[kuruman]

I don't understand what you are doing here: $$dI=\int_{x=0}^{x=R}{\frac{I}{\pi R^2/2}\times\frac{d\theta}{2\pi}\times 2\pi x dx}.$$ The semicircular ring of radius $R$ is very thin and there should be no $\pi R^2$ term to consider. Assume that the current is uniformly distributed over the ring. An arc $ds=R ~d\theta$ has current $$dI=\frac{I}{\pi R}ds=\frac{1}{\pi}I~d\theta.$$ Consider that as an infinitely long wire, find the two components of its contribution $dB_x$ and $dB_y$ at the center and integrate over $\theta$. The vertical component is expected to integrate to zero by symmetry. That's how one adds vectors.

 

[Aurelius120]

I don't understand what you are doing here: $$dI=\int_{x=0}^{x=R}{\frac{I}{\pi R^2/2}\times\frac{d\theta}{2\pi}\times 2\pi x dx}.$$ The semicircular ring of radius $R$ is very thin and there should be no $\pi R^2$ term to consider.

The wire has semi-circular cross section. Therefore it has current distributed throughout its volume . It is a half cylinder with area $\pi R^2/2$ so current density is calculate by dividing current by area. So what is wrong with how I did it?
Why are you dividing current with circumference to find current density?

 

[haruspex]

It is a half cylinder

The text says "semicircular ring", so just an arc of negligible width.

 

[Aurelius120]

The text says "semicircular ring", so just an arc of negligible width.

So how does that look like?

 

[Steve4Physics]

@Aurelius120 (just in case you're still not clear) based on the Post #1 diagram the conductor is shaped like this

(but infinitely long and made of material of negligible thickness).

The use of the terms 'wire'and 'ring' in the question, is not ideal IMO.