Problem 15.3 Challenge

จาก \((π - θ) + 2α = π\)

\(\alpha =\frac{\theta}{2}\)

จาก \(α + β = \frac{\pi}{2}\)

\(\beta =\frac{\pi}{2}-\alpha =\frac{\pi}{2}-\frac{\theta}{2}\)

จาก \(d\vec{B} = \frac{\mu _{0}}{4\pi}\frac{Id\vec{s}\times r̂}{r^{2}}\)

k̂

\(d\vec{B} =\frac{\mu _{0}I}{4\pi r^{2}}(\sin \beta ds)\)

k̂

\(d\vec{B} =\frac{\mu _{0}I}{4\pi r^{2}}(\sin \beta ds)\)

k̂

\(d\vec{B} =\frac{\mu _{0}I}{4\pi (2Rcos\alpha )^{2}}(\sin \beta ds); \cos \alpha =\frac{r}{2R}\)

k̂

\(d\vec{B} =\frac{\mu _{0}I}{4\pi (2Rcos\alpha )^{2}}\sin (\frac{\pi}{2}-\frac{\theta}{2})ds;\beta =\frac{\pi}{2}-\frac{\theta}{2}\)

k̂

\(d\vec{B} =\frac{\mu _{0}I}{4\pi (2Rcos\alpha )^{2}}\cos (\frac{\theta}{2})ds\)

k̂

\(d\vec{B} =\frac{\mu _{0}I}{16\pi R^{2}\cos ^{2}(\frac{\theta}{2})}\cos (\frac{\theta}{2})ds; \alpha =\frac{\theta}{2}\)

k̂

\(d\vec{B} =\frac{\mu _{0}I}{16\pi R^{2}}\sec (\frac{\theta}{2})ds\)

k̂

\(d\vec{B} =\frac{\mu _{0}I}{16\pi R^{2}}\sec (\frac{\theta}{2})Rd\theta ; ds = Rd\theta\)

k̂

\(d\vec{B} =\frac{\mu _{0}I}{16\pi R}\sec (\frac{\theta}{2})d\theta\)

\(^{\pi /2}k̂\)

\(\vec{B} =\frac{\mu _{0}I}{16\pi R}\int \sec (\frac{\theta}{2})d\theta\)

\(_{-\pi /2}\)

\(_{\pi /2}\)

k̂

\(\vec{B} =\frac{\mu _{0}I}{8\pi R}\ln (\sec (\frac{\theta}{2})+\tan (\frac{\theta}{2}))\)

\(_{-\pi /2}\)

k̂

\(\vec{B} =\frac{\mu _{0}I}{8\pi R}\ln (\frac{2 + 1}{2 - 1}) \;\blacksquare\)