Chapter 27 Problem 69

Problem

A sort of “projectile launcher”

A large current moves in a closed loop composed of fixed rails, and there is a very light, almost frictionless bar touching the rails. A 1.8 T magnetic field is perpendicular to the plane of the circuit. If the rails are a distance $d$ = 24 cm apart, and the bar has a mass of 1.5 g, what constant current flow is needed to accelerate the bar from rest to 25 m/s in a distance of 1.0 m? In what direction must the field point?

solution

The accelerating force on the bar is due to the magnetic force on the current. If the current is constant, the magnetic force will be constant, and so constant acceleration kinematics can be used.

$v^{2}=v_{0}^{2}+2a\Delta x\rightarrow a={\frac {v^{2}-0}{2\Delta x}}={\frac {v^{2}}{2\Delta x}}$

$F_{\textrm {net}}=ma=IdB\rightarrow I={\frac {ma}{dB}}={\frac {m{\frac {v^{2}}{2\Delta x}}}{dB}}={\frac {mv^{2}}{dB2\Delta x}}$ $={\frac {(1.5\times 10^{-3}{\textrm {kg}})(25{\textrm {m/s}})^{2}}{2(1.0{\textrm {m}})(0.24{\textrm {m}})(1.8{\textrm {T}})}}=1.1{\textrm {A}}$

Right hand rule: Magnetic field is pointing down.