Answer
1.2 amps, and the magnetic field points downward.
Work Step by Step
If the current is constant, the magnetic force will be constant. Use the equations for constant acceleration to find the desired acceleration.
$$v^2=v_o^2+2a\Delta x$$
$$a=\frac{v^2- v_o^2}{2\Delta x }=\frac{(28m/s)^2-0^2}{2(1.0m)}=392m/s^2$$
The current is perpendicular to the magnetic field, so use equation 20–2 to find the force.
$$F_{net}=I\mathcal{l}B=ma$$
Solve for the required current.
$$I=\frac{ma}{\mathcal{l}B }=\frac{(1.5\times10^{-3}kg)( 392m/s^2)}{(0.28m)(1.7T)}=1.2A$$
Use the right-hand rule to find the direction of the magnetic field. The magnetic field must point downward so that the force on the projectile is in the desired direction (shown in Figure 20–71).
