Problem 15.2 Challenge

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

îĵk̂

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

\(00ds\)

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

î

\(d\vec{B} =\frac{\mu _{0}I}{4\pi r^{2}}\cos \theta ds +\frac{\mu _{0}I}{4\pi r^{2}}\sin \theta ds ˄\perp i\)

\(^{2\pi R}î\)

\(\vec{B} =\frac{\mu _{0}Icos\theta}{4\pi r^{2}}\int ds\)

\(_{0}\)

\(_{2\pi R}\)

î

\(\vec{B} =\frac{\mu _{0}Icos\theta}{4\pi r^{2}}s\)

\(_{0}\)

î

\(\vec{B} =\frac{\mu _{0}I(2\pi R)\cos \theta}{4\pi r^{2}}\)

î

\(\vec{B} =\frac{\mu _{0}I(2\pi R)}{4\pi r^{2}}(\frac{R}{r})\)

î

\(\vec{B} =\frac{\mu _{0}IR^{2}}{2r^{3}}\)

î

\(\vec{B} =\frac{\mu _{0}IR^{2}}{2(x^{2} + R^{2})^{3/2}}\;\blacksquare\)

พิจารณากรณีจุด P อยู่ที่จุดศูนย์กลางของเส้นลวดวงกลม

จะได้ \(x = 0\)

î

\(\therefore \vec{B} =\frac{\mu _{0}I}{2R}\)