Problem 15.1 Challenge

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

îĵk̂

\(d\vec{B} =\frac{\mu _{0}I}{4\pi (acosec\theta )^{2}}\)

\(dx00\)

\(\cos \theta \sin \theta 0\)

k̂

\(d\vec{B} =\frac{\mu _{0}Isin\theta dx}{4\pi (acosec\theta )^{2}}\)

k̂

\(\vec{B} =\frac{\mu _{0}I}{4\pi a^{2}}\int \sin ^{3}\theta dx\)

จาก \(cotθ = \frac{-x}{a}\)

\(x = -acot\theta\)

\(\frac{dx}{d\theta}= cosec^{2}\theta\)

\(\therefore dx = cosec^{2}\theta d\theta\)

\(^{\theta _{2}}k̂\)

จะได้ \(\vec{B} = \frac{\mu _{0}I}{4\pi a^{2}}\int \sin ^{3}\theta (acosec^{2}\theta d\theta )\)

\(_{\theta _{1}}\)

\(^{\theta _{2}}k̂\)

\(\vec{B} =\frac{\mu _{0}I}{4\pi a}\int \sin \theta d\theta\)

\(_{\theta _{1}}\)

\(_{\theta _{2}}\)

k̂

\(\vec{B} =\frac{\mu _{0}I}{4\pi a}(-\cos \theta )\)

\(_{\theta _{1}}\)

k̂

\(\vec{B} =\frac{\mu _{0}I}{4\pi a}(\cos \theta _{1} - \cos \theta _{2}) \;\blacksquare\)

พิจารณากรณีเส้นลวดยาวมากๆ

จะได้ \(θ_{1} \to 0\) และ \(θ_{2} \to π\)

k̂

\(\therefore \vec{B} =\frac{\mu _{0}I}{4\pi a}(\cos 0 - \cos \pi )\)

k̂

\(\vec{B} =\frac{\mu _{0}I}{2\pi a}\)