Question

A 1.00 (m) metal bar that has a mass of 0.490 (kg) is initially pinned in place on an incline 65.0 above the horizontal plane. There is a 0.850 (T) magnetic field directed upward in the region around the bar, and metal contacts along the incline allow a current through the bar from D to C. The coefficient of static friction between bar and contacts is 0.2. What is the minimum magnitude of the current that must pass through the bar to keep it from moving once it is released?

Answer 1

Answer:

the minimum magnitude of the current that must pass through the bar to keep it from moving once it is released is 7.69 A

Explanation:

From the question; the normal force acting on the metal bar can be written as:

N = Fsin θ + mg cos θ

But F = ILB

However, the expression for the frictional force  along the incline can be written as:

$F_f = mg sin \theta - F cos \theta$

But $f = \mu N$

So; we can rewrite our formula as:

$f = \mu ( F sin \theta + mg cos \theta)$

Thus the expression for the minimum magnitude and direction of current which passes through the bar can be expressed as:

$\mu (ILB) sin \theta + mg cos \theta = mg sin \theta - (ILB) cos \theta$             since F = ILB

Making I the subject of the formula ; we have:

$I = \frac{mg ( sin \theta - \mu cos \theta )}{LB ( \mu sin \theta + cos \theta )}$

Given that;

mass   m = 0.490 kg

acceleration due to gravity = 9.8

the angle of incline θ = 65°

coefficient of static friction between bar and contacts ($\mu$) =  0.2

magnetic field (B) = 0.850 (T)

Length of the metal bar = 1.00 m

Therefore:

$I = \frac{0.490*9.8 ( sin 65^0 - (0.2) cos (65^0) )}{ 1*(0.850)((0.2) sin 65^0 + cos 65^0 ))}$

I = 7.69 A

Therefore, the minimum magnitude of the current that must pass through the bar to keep it from moving once it is released is 7.69 A