Induced Current in a Loop: Third Problem of Cluster, Exercises of Electrical Engineering

Download Induced Current in a Loop: Third Problem of Cluster and more Exercises Electrical Engineering in PDF only on Docsity! 108. (Third problem of Cluster) (a) , (b) and (c) The area enclosed by the loop is that of a rectangle with one side (x) expanding. With B0 = 0.200 T and ξ = 0.050 T/s (the rate of field increase), we have Φ = BA = (B0 + ξt) (Lx) = B0Lvt + ξLvt2 where x = vt has been used. Thus, from Faraday’s and Ohm’s laws, the induced current is i = E R = B0Lv R + 2 ξLv R t and is counterclockwise (to produce field in the loop’s interior pointing out of the page, “fighting” the increasing inward pointed flux due to the applied field). Therefore, the current at t = 0 is B0Lv/R = 0.144 A. And its value at t = 1.00 s is (B0 + 2ξ)Lv/R = 0.216 A. docsity.com