Question

This problem explores how a current-carrying wire can be accelerated by a magnetic field. You will use the ideas of magnetic flux and the EMF due to change of fluxthrough a loop. Note that there is an involved follow-up part that will be shown once you have found the answer to Part B.

What is the acceleration $a_r(t)$ of the rod? Take to be the mass of the rod.Express your answer as a function of , , the velocity of the rod $v_r(t)$ , , , and the mass of the rod $a_r(t)$ = ?

Answer 1

The magnetic flux of the loop is, \(\phi_{B}=B L x(t)\) The induced \(e m f\) is $$ \begin{aligned} \varepsilon_{i v d} &=\frac{-d \phi}{d t} \\ &=\frac{-B L x(t)}{d t} \\ &=-B L v(t) \end{aligned} $$ The net \(e m f\) in the circuit is, $$ \begin{aligned} \varepsilon_{\text {net}} &=V+\varepsilon_{\text {ivd}} \\ &=V-B L v(t) \end{aligned} $$ The current in the circuit is, The acceleration is $$ \begin{aligned} I_{\max } &=\frac{\varepsilon_{\text {net}}}{R} & a(t) &=\frac{F(t)}{m} \\ &=\frac{V-B L v(t)}{R} & &=\frac{\left(\frac{V-B L v(t)}{R}\right) L B}{m} \end{aligned} $$ \(\begin{aligned} F(t) &=B I L \\ &=\left[\frac{V-B L v(t)}{R}\right] B \cdot L \end{aligned}=\frac{[V-B L v(t)] L B}{R \cdot m}\)