A rod of mass 0.720 kg and radius 6.00 cm rests on two parallel rails (see figure below) that are d = 12.0 cm apart and L = 45.0 cm long. The rod carries a current of I = 66.0 A in the direction shown and rolls along the rails without slipping. A uniform magnetic field of magnitude 0.230 T is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails? (Assume that the rod is of uniform density.)

The problem involves the application of the laws of physics, specifically the principles of electromagnetism and mechanics. The magnetic force acting on the rod can be calculated using the formula F = BIL, where B is the magnetic field, I is the current, and L is the length of the rod. This force will cause the rod to accelerate along the rails, and we can use the equations of motion to find the final speed of the rod.

Step 1: Calculate the magnetic force acting on the rod. F = BIL = (0.230 T)(66.0 A)(0.45 m) = 6.831 N

Step 2: Calculate the acceleration of the rod. The force acting on the rod will cause it to accelerate. We can find the acceleration using Newton's second law, F = ma. a = F/m = 6.831 N / 0.720 kg = 9.49 m/s²

Step 3: Calculate the final speed of the rod. We can find the final speed using the equation of motion v² = u² + 2as, where u is the initial speed, a is the acceleration, and s is the distance. The rod starts from rest, so u = 0. v² = 0 + 2(9.49 m/s²)(0.45 m) = 8.54 m²/s² Taking the square root of both sides gives v = 2.92 m/s.

So, the speed of the rod as it leaves the rails is 2.92 m/s.

The problem involves the application of the law of conservation of energy and the force on a current-carrying rod in a magnetic field.

The magnetic force on the rod is given by F = ILB, where I is the current, L is the length of the rod, and B is the magnetic field. This force will cause the rod to accelerate along the rails.

The work done by this force as the rod moves along the rails is given by W = Fd, where d is the distance the rod travels. This work goes into the kinetic energy of the rod, which is given by KE = 1/2m*v^2, where m is the mass of the rod and v is its speed.

Setting the work equal to the kinetic energy and solving for v gives:

v = sqrt((2Fd)/m)

Substituting the expression for F from above gives:

v = sqrt((2ILBd)/m)

Substituting the given values gives:

v = sqrt((266.0 A0.45 m0.230 T0.12 m)/(0.720 kg))

v = sqrt((266.00.450.2300.12)/0.720)

v = sqrt(0.8712)

v = 0.933 m/s

So the speed of the rod as it leaves the rails is 0.933 m/s.