workout 31 - HPHY 253 Workout 31 Faraday&s Law of...

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Unformatted text preview: HPHY 253 Workout 31 Faraday&s Law of Electromagnetic Induction Dr. A. E. Bak Name 1. Serway & Jewett Problem 31 : 8 . A rectangular wire loop of length L and width w and an in¡nitely long, thin straight wire carrying a current I both lie on a tabletop as shown in Figure P 31 : 8 . Take the xy plane as the plane of the tabletop. More speci&cally, take the x axis along the in&nitely long wire, with the positive direction as that of its current. Moreover, take the wire loop to lie between x = 0 and x = L and between y = h and y = h + w . (a) Determine the magnetic ¢ux through the loop due to the current I . Let surface S be the rectangular region bounded by the wire loop. The magnetic &eld on surface S , which is produced by the current I , is given by B = & I e z 2 ¡y : The area da of a rectangular strip of S , lying between y and y + dy , is given by da = Ldy ; having associated area vector da = da e z = ( Ldy ) e z : The magnetic ¡ux d & through this in&nitesimal area is given by d & = B & da = & & I e z 2 ¡y ¡ & ( Ldy e z ) = & IL 2 ¡ dy y : Therefore, we have & = Z S d & = & IL 2 ¡ Z h + w h dy y = & IL 2 ¡ ln & h + w h ¡ as the magnetic ¡ux through surface S . (b) Suppose that the current varies with time t according to the relation I = a + bt , where a and b are constants. Determine the induced emf in the loop if b = 10 : A=s , h = 1 : 00 cm , w = 10 : cm , and L = 100 cm . On making use of Faraday¢s law, we obtain E = ¡ d & dt = ¡ & L 2 ¡ dI dt ln & h + w h ¡ = ¡ & Lb 2 ¡ ln & h + w h ¡ as the induced emf in the wire loop. We &nd that E = ¡ 4 : 80 ¢ 10 & 6 V = ¡ 4 : 80 & V for the given values of b , h , w , and L . (c) What is the direction of the induced current in the rectangular loop? Let CCW and CW re- spectively denote £counterclockwise¢and £clockwise.¢ There is a right-hand rule that relates the respective directions of the da vectors of surface S and the dl vectors of its boundary. According to this rule, we have da points into page = ) dl points CW along boundary : Because E < , the direction of the induced current is opposite that of dl . That is, the induced current is CCW along the boundary. 2. Serway & Jewett Problem 31 : 13 . Find the current through section PQ of length a = 65 : cm in Figure P 31 : 13 . The circuit is located in a magnetic ¡eld whose magnitude varies with time according to the relation B = 1 : 00 t ¢ 10 & 3 , where all quantities here are measured in the appropriate SI units. Assume that the resistance per length of wire is : 100 ¡ =m . 1 & Determining the induced emfs: Let surface S 1 .be the square region bounded by the right-hand loop. For this surface, we have B = B ¡ ; A = a 2 ¡ ; and so the magnetic &ux through S 1 is & 1 = B ¢ A = Ba 2 = 4 : 225 t £ 10 & 4 : It follows that E 1 = ¤ d & 1 dt = ¤ 4 : 225 £ 10 & 4 V = ¤ : 4225 m V : There is a right-hand rule that relates the respective directions of the...
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This note was uploaded on 05/01/2011 for the course HPHY 253 taught by Professor Bak during the Spring '09 term at Morehouse.

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workout 31 - HPHY 253 Workout 31 Faraday&s Law of...

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