Fall 2003 4 Em

A metallic "horseshoe-shaped" loop of width $a$ is attached to a sliding wire, as shown in the figure. A uniform magnetic field is perpendicular to the plane of the system. If the wire has mass $M$ and resistance $R$, how far will it travel if it is given an initial speed $v_0$?

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When the wire moves there will be an EMF produced, which creates a current. The magnetic field provides a force on the current slowing it down. Mathematically, this is $EMF=-\frac{1}{c}\frac{d\Phi}{dt}$, where $\Phi=BA$. $B$ is the magnitude of the magnetic field inside the loop and $A$ is the area of the loop. $B$ is constant and $A=K+x(t)a$, where $K$ is the initial area of the closed loop and $x(t)$ is the position of the wire from it's initial position. Then $\frac{d\Phi}{dt} = B\frac{dA}{dt}=Bav(t)$. We therefore get

(1)
\begin{align} EMF=-\frac{Bav(t)}{c}=IR \end{align}

The fore on the wire is

(2)
\begin{align} F=\frac{IaB}{c} = -\frac{B^2 a^2}{c^2}\frac{v(t)}{R}=Ma \end{align}

We can solve this equation for $x(t)$ as a function of time, with initial conditions $x(0)=0$ and $v(0)=v_0$, giving

(3)
\begin{align} x(t) = -\frac{Rc^2}{B^2 a^2} v_0 \exp \left( -\frac{B^2 a^2}{R c^2} t \right) + \frac{R c^2}{B^2 a^2} v_0 \end{align}

The distance the loop travels is

(4)
\begin{align} x(\infty) = \frac{R c^2}{B^2 a^2} v_0 \end{align}