1 point A rod of length 10 cm and mass m 50 g lies on a pla

(1 point) A rod of length 10 cm and mass m = 50 g lies on a plane inclined at = 28 to the horizontal, as shown in the figure above. A current enters and leaves the rod via light flexible wires which we ignore. For what current magnitude will the rod be in equilibrium in a magnetic field B 0.3/T? Ignore friction.

Solution

The current is in the -z direction and B is in the y direction, so the magnetic force is;
F = (iL)(-k) X B(j) = iLB(i) , the x-direction.

So the forces on the rod are;
weight, straight down
Normal, perpendicular to the plane
Magnetic, to the right.

Taking components parallel and perpendicular to the plane gives the equilibrium eqs;
mgSin(28) = iLBCos(28)
N = mgCos(28) + iLBSin(28)

Use the top eq to find the answer
i = [mgTan(28)]/LB

i = (50 *10^-3kg )(9.8m/s²) (0.53170) / (10*10^-2m)(0.3T)

i= 8.684A