假設回路中電阻恒為 R ,兩個小棒質素均為 m ,兩小棒長度均為 l ,取 v_0 方向為正
\begin{split} E &=Bl(v_1 - v_2) \end{split}
I= \frac{E}{R} = \frac{Bl}{R}(v_1 - v_2)
安培力大小 F = BIl = \frac{Bl^2}{R}(v_1 - v_2)
由牛頓第二定律知,對於棒 ab 的加速度為 -\frac{F}{m} ,對於棒 cd 則為 \frac{F}{m}
也即
\left\{ \begin{aligned} \frac{\mathrm dv_1}{\mathrm dt} &=-\frac{Bl^2}{Rm}(v_1 - v_2)\cr \frac{\mathrm dv_2}{\mathrm dt} &=\frac{Bl^2}{Rm}(v_1 - v_2)\cr \end{aligned} \right.
記 a = \frac{Bl^2}{Rm} 得原方程式組為
\left\{ \begin{aligned} \frac{\mathrm dv_1}{\mathrm dt} &=a(v_2 - v_1)\cr \frac{\mathrm dv_2}{\mathrm dt} &=a(v_1 - v_2)\cr \end{aligned} \right.
由方程式組的上式得
v_2 = \frac{1}{a}\cdot \frac{\mathrm d v_1}{\mathrm dt} + v_1
兩邊求微分得
\frac{\mathrm dv_2}{\mathrm dt} = \frac{1}{a}\cdot \frac{\mathrm d^2 v_1}{\mathrm dt^2} + \frac{\mathrm dv_1}{\mathrm dt}
將這二式代入到方程式組的下式得
\frac{1}{a}\cdot \frac{\mathrm d^2 v_1}{\mathrm dt^2} + \frac{\mathrm dv_1}{\mathrm dt} =a\left(v_1 - \frac{1}{a}\cdot \frac{\mathrm d v_1}{\mathrm dt} - v_1\right)
即得
\frac{1}{a}\cdot \frac{\mathrm d^2 v_1}{\mathrm dt^2} + 2\cdot\frac{\mathrm dv_1}{\mathrm dt} =0
解得
v_1 =C_1 + C_2 e^{-2at} .
代到方程式組下式,並解之得
\begin{split} v_2&=C_1 - C_2e^{-2at}+Ce^{-at}\cr \end{split}
由 t = 0 時, v_1 = v_0 , v_2 = 0 得
\left\{ \begin{aligned} &C_1 + C_2 = v_0 \cr &C_1-C_2+C = 0 \end{aligned} \right.
當 t\to +\infty , v_1 \to C_1 , v_2 \to C_1 .
又由於整個過程,二小棒動量守恒。
故 mv_0 = mC_1 + mC_1
也即 C_1 = \frac{v_0}{2} .
進而解得
\left\{ \begin{aligned} C_1 &=\frac{v_0}{2}\cr C_2 &= \frac{v_0}{2}\cr C &= 0 \end{aligned} \right.
也即
\left\{\begin{aligned} v_1 &=\frac{v_0}{2}\left(1 + e^{-2\frac{Bl^2}{Rm}t}\right)\cr v_2 &=\frac{v_0}{2}\left(1 - e^{-2\frac{Bl^2}{Rm}t}\right)\cr \end{aligned}\right.
由此可得
\begin{split} I &= \frac{Bl}{R}(v_1 - v_2)\cr &=\frac{Bl}{R}\cdot v_0e^{-2\frac{Bl^2}{Rm}t}\cr \end{split}