Analysis of an Electromagnetic Problem: Current and Induced EMF, Exercises of Electrical Engineering

Download Analysis of an Electromagnetic Problem: Current and Induced EMF and more Exercises Electrical Engineering in PDF only on Docsity! 107. (Second problem of Cluster) (a) With L = 0.50 m and R = 5.00 Ω, we combine Ohm’s and Faraday’s laws, so that the current magnitude is i = |E| R = BLv R = 0.240 A . The direction is counterclockwise, as explained in the solution to the previous problem. (b) The area in the loop is A = 12 (L0 + L)x where x = vt and L0 = 0.300 m. But the value of L depends on the distance from the resistor x: L = 30 cm + ( 20 cm 1 m ) x = L0 + 0.200(vt) where x = vt has been used. Therefore, the area becomes A = L0 vt+ 0.100 v2t2 . The induced emf is, from Faraday’s law, E = dΦ dt = B dA dt = B ( L0 v + 2(0.100)v2t ) and the induced current is i = E R = 0.144 + 1.152t in SI units and is counterclockwise (for reasons given in previous solution). docsity.com