Question

1) Figure 1 shows an infinity long wire with current I in +x direction. A triangular loop with the each side equal L and the total resistance of R is placed initially at distance of do from the wire as shown in the figure. a) Find induced current (including the direction) if the loop is moving with constant velocity of 3 = Vol. b) Find induced current (including the direction) if the loop is moving with constant velocity of v = vol. do Fig. 1 Problem (1)

Answer 1

Я b. L Lok T I ly B- Moguthe field at y distance llo any fluxe through the strip of dy thickness do = BdA = B x body b. length of strip = L-214-do) cot too A Cot 60°-BC AC Cot 600 - BC A Ty do hence =) BC = (y-do) cot 60° b=2-2(y-do) cot 60° =) do = B x bdy = Mo I {1-244-defentseady day lo 274 { 2 - 2 (y-doję dy MOI 2x 2 (1 do jędy y 3

va Page No: Total flux through the triangle b= (do = {MI 2 1 1 2 3 (1 - do ), dy limit of inzigration do to do + 13 / (attitude of equilateral tiangle Mo I se te dhome) dy Mo I buy - 2 y + 200 ling 27 lol I to do the 2 do (do to - do) dot B L ado 13 lu =) & = do I Į ln (1+ B ) - L+² L+ 2do 2x 3 206

(a) Induced current I I= 1 2 2 tot (Leu (1+13L Ratan 2do) HBL = I= -llo I IL QARCHBL de do) -0 zdo + zelor V3 2007) 2do de = I= llo I 8 -3 2² L %ddol QARI 2do + Bldo do + BLS at motion = Vo that means in yolirection hence Voz d(do) = I= MIL I SBL bado 1 1 2 0 4 3L + 20 gr - Is Mo IL VO S B2+2 do q Zardo [VBL +2 do lo IL Vo cardo

direchow of induced current wëll be so that it can resist the change en flux. As coil moves in ty direction B decreases and hence flux. So current in loop will be so that it can increase fleex and hence magnetic field Thus, for this direction will be Corinter clockwise, viewed from top (b) In this case do will be constant hence dedo) 20 - 0 -) I