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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 = 1 2 ( L 0 + L ) x where x = vt and L 0 =0 . 300 m. But the value of L depends on the distance from the resistor x : L =3 0 c m + µ 20 cm 1m ¶ x = L 0
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