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A sort of projectile launcher is built in the following way: A large current moves in a closed loop composed of fixed rails, a power supply and a very light, almost frictionless bar touching the rails. A magnetic field is perpendicular to the plane of the circuit. If the bar has length $\SI{18}{cm}$, a mass of $\SI{1.3}{g}$, and is placed in a field of $\SI{1.7}{T}$, what constant current flow is needed, to accelerate the bar from rest to $\SI{27}{\meter\per\second}$ in a distance of $\SI{1.2}{m}$? In what direction must the magnetic field point? $\star$
\begin{empheq}[box=\Gegeben]{align} \ell &= \SI{18}{cm} = \SI{0.18}{m} \\ m &= \SI{1.3}{g} = \SI{1.3e3}{kg}\\ B &= \SI{1.7}{T} \\ v &= \SI{27}{\meter\per\second}\\ s &= \SI{1.2}{m} \end{empheq} \begin{empheq}[box=\Gesucht]{align} \text{Electric current (flow), } [I]=\si{A} \end{empheq} The acceleration needed, to reach the given velocity in the given distance, is: \begin{align} a &= \frac{v^2}{2s}\\ &= \SI{303.75}{\meter\per\second\squared} \end{align} The force required to accelerate an object with the given mass to the above calculated value, is: \begin{align} F&= ma = m \frac{v^2}{2s} = \frac{mv^2}{2s}\\ &= \SI{0.395}{N} \end{align} Hence, the current in the wire must be: \begin{align} I &= \frac{F}{B\ell} = \frac{\frac{mv^2}{2s}}{B\ell} = \frac{mv^2}{2s\ell B}\\ &= \SI{1.29}{A} \end{align} \begin{empheq}[box=\Lsgbox]{align} I &= \frac{mv^2}{2s\ell B}\\ &= \SI{1.29}{A} \end{empheq}
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