- #1
Aurelius120
- 251
- 24
- 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
$$dB=\int^{x=R}_{x=0}{\frac{\mu_○}{2\pi x}\times\frac{I}{\pi R^2/2}\times\frac{d\theta}{2\pi}\times 2\pi x dx}$$
$$\implies dB=\frac{\mu_○IR}{\pi^2R^2}d\theta$$
![1000016289.jpg 1000016289.jpg](https://www.physicsforums.com/data/attachments/324/324286-cdd092cf9ed959000ee0cc0100143ee6.jpg)
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?