Problem 15.54 Challenge
หัวข้อ: แรงแม่เหล็กระหว่างตัวนำไฟฟ้าสองตัวที่ขนานกัน (The Magnetic Force Between Two Parallel Conductors)
\(\vec{F} = -\frac{\mu _{0}I_{1}I_{2}hw}{2\pi \ell (\ell + w)}\;\blacksquare\)
ดูวิธีทำ
พิจารณาเส้นส่วนที่ \(1\)
ĵk̂î
จาก \(\vec{F}_{1} = I_{2}(h )\times (- \frac{\mu _{0}I_{1}}{2\pi \ell}) = -\frac{\mu _{0}I_{1}I_{2}h}{2\pi \ell}\)
พิจารณาเส้นส่วนที่ \(2\)
îk̂
จาก \(d\vec{F}_{2} = I_{2} (dx_{2} )\times (-\frac{\mu _{0}I_{1}}{2\pi x_{2}})\)
ĵ
\(d\vec{F}_{2} =\frac{\mu _{0}I_{1}I_{2}dx_{2}}{2\pi x_{2}}\)
\(^{\ell + w}ĵ\)
\(\vec{F}_{2} =\frac{\mu _{0}I_{1}I_{2}}{2\pi}\int \frac{dx_{2}}{x_{2}}\)
\(_{\ell}\)
ĵ
\(\vec{F}_{2} =\frac{\mu _{0}I_{1}I_{2}}{2\pi}\ln (\frac{\ell + w}{w})\)
พิจารณาเส้นส่วนที่ \(3\)
ĵk̂î
จาก \(\vec{F}_{3} = I_{2}(-h )\times (-\frac{\mu _{0}I_{1}}{2\pi (\ell + w)}) =\frac{\mu _{0}I_{1}I_{2}h}{2\pi (\ell + w)}\)
พิจารณาเส้นส่วนที่ \(4\)
îk̂
จาก \(d\vec{F}_{4} = I_{2} (-dx_{4} )\times (-\frac{\mu _{0}I_{1}}{2\pi x_{4}})\)
ĵ
\(d\vec{F}_{4} = -\frac{\mu _{0}I_{1}I_{2}dx_{4}}{2\pi x_{4}}\)
\(^{\ell + w}ĵ\)
\(\vec{F}_{4} = -\frac{\mu _{0}I_{1}I_{2}}{2\pi}\int \frac{dx_{4}}{x_{4}}\)
\(_{\ell}\)
ĵ
\(\vec{F}_{4} = -\frac{\mu _{0}I_{1}I_{2}}{2\pi}\ln (\frac{\ell + w}{w})\)
\(\therefore \vec{F} = \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3} + \vec{F}_{4}\)
îĵ
\(\vec{F} = -\frac{\mu _{0}I_{1}I_{2}h}{2\pi \ell}+\frac{\mu _{0}I_{1}I_{2}}{2\pi}\ln (\frac{\ell + w}{w})\)
îĵ
\(+\frac{\mu _{0}I_{1}I_{2}h}{2\pi (\ell + w)}-\frac{\mu _{0}I_{1}I_{2}}{2\pi}\ln (\frac{\ell + w}{w})\)
î
\(\vec{F} =\frac{\mu _{0}I_{1}I_{2}h}{2\pi}(\frac{1}{\ell + w}-\frac{1}{\ell})\)
î
\(\vec{F} = -\frac{\mu _{0}I_{1}I_{2}hw}{2\pi \ell (\ell + w)}\;\blacksquare\)