Question

please answer all parts a,b,c,d

(17%) Problem 2: A wire, bent into a rectangle with sides a = 0.075 m and b = 0,045 m, is in a magnetic field B directed perpendicularly to the face of the wire, as shown. The perpendicular component is a function of time as B(t) = Asinot), where A = 0.45 T. 6 = 9 rad/s. In this problem, take the normal vector to the surface of the loop to be parallel to the magnetic field. Otheexpertta.com 25% Part(a) Express the magnetic flux going through the wire, in terms of a, b, 4.0, and I Grade Summary Deductions Potential 1005 HOME cos(a) cos(0) sin(at) cos(ot) sin(a) sin() cos(0) sin(6) sin(0) b ( 7 8 9 AAL 4 5 6 1 2 3 Submissions Attempts remaining: ( O per attempt) detailed view d V VODACO CHER Submit Hint Feedback I give up! Hints: O deduction per hint. Hints remaining: 1 Feedback: O deduction per feedback A 25% Part (b) Express the the emf, 8. induced in the loop in terms of the derivative of 25% Part (c) Find the emf, induced in the loop in terms of A, w, a, and b. 25% Part (d) Calculate the magnitude of the emf at 1 = 8.6 s in V.

Answer 1

This problem is from topic electromagnetic Induction.

(a) flux (4 is gine we by - know magnetic $ = Ē Ā where u is À is magnetic area field vector Vector $ = B. À هي 268 = 6 و چیاى من و angle between and A So, Hea = 0º caso = d = 18 1 A (caso) $ = [A cinct)] [ab] So 6= Aab senat to Faradays we low know of E = According emf -do Here emf flux negative will through sign indicates induced oppose the change in the loop.

{- - do at - = absin (cat) E = - (Aab w) cos(at) SO E = Aabww cos(at) As we are concerned with absolute value of emf so we well omit negative sign. (d) E= magnitude of Given emf = Aabo cos(at) A= 0.45T a= 0.075 m b= 0.045m =د 9 /اعت t = E = 0.00571 V E 8.6 s = 5.71x10-3 v T E= 5.71x10-3 VT