For a constant magnetic field, the magnetic flux through a loop is given by the area of the loop times the magnetic field times the cosine of the angle between the field and the normal. For a square loop, the area is (2.00 m)2 = 4.00 m2. This means that the flux is (4.00 m2)(0.300 T)(cos 50o) = .77 T M2.
For a multiple coil loop, the flux is given by NAB times the cosine of the angle between the magnetic field and the normal of the coil. For an initial alignment with the field, the flux is (500)(3.14)(.075 m)2B = 8.83 B m2. If the coil is flipped, then the change in the flux is twice this, or 17.7 B m2. If this flux change occurs in .00277 s, then the rate of flux change is 6390 B m2/s. This rate of change in the flux is equal to the induced voltage. This means that 6390 B m2/s = 0.166 V. This gives B = (0.166 V)/(6390 m2/s) = 2.6 x 10-5 T.
From Figure 20.35, we can see that the resistor, rails, and slidable bar form a rectangle through which a magnetic field is present. The flux through this rectangle is given by BA = (2.5 T)(1.2 m)(x) = (3.0 T m) x where x is the horizontal displacement of the rail. Since the magnetic field is not changing, the rate of change of the flux is, therefore, given by (3.0 Tm) (change in x)/(change in time) = (3.0 Tm) v where v is the velocity of the rail. This change in flux is equal to the induced voltage in the circuit, which is given by V = IR = (.50 A)(6.0 ohms) = 3.0 V. Equating these two expressions, we get (3.0 T m) v = 3.0 V, which gives v = 1 m/s.
We know that the time constant for an RL circuit is t = L/R. For this problem, we are given that t = 600 microseconds. However, we need to find R before we can find L. When the current is just beginning to flow, the inductor acts as a back emf to stop the flow of the current. As the current builds up, the change in the current decreases, decreasing the amount of the back emf. After a while, the current will reach its maximum, and there will be no voltage across the inductor. Therefore, the voltage across the resistor will be equal to the battery's voltage at this point, i.e. V = IR. This gives R = V/I = 6.0 V/.3 A = 20 ohms. Using the 600 microsecond time constant, we get L = Rt = (20 ohms)(.0006 s) = 1.2 x 10-2 H.
If you find an error in these, or just have a comment, send an e-mail message to jpratte@kennesaw.edu