Question

The figure below shows a rectangular loop of wire that is 0.300 m wide, is 1.50 m long, and lies in the vertical plane which is perpendicular to a region that has a uniform magnetic field. The magnitude of the uniform magnetic field is 0.400 T and the direction of the magnetic field is into the page. The portion of the loop not in the magnetic field is 0.100 m long. The resistance of the loap is 0.200 2 and its mass is 50.0 g. The loop is released from rest at t = 0. (Let dawn be the positive y-direction.) (a) What are the magnitude and direction of the induced current when the luop has a downward speed v? (Use the following as necessary: 8, R, V, and w for the width of the loop. Do not substitute numerical values; use variables only.) megnitude - direction dockwise b) What is the force that acts on the loop as a result of the current? (Use the following as necessary: B, R, V, and w for the width of the loop. Do not substitute numerical values; use variables only) magnitude - BV direction ward ! (c) What is the net force acting on the loup? (Use the following as necessary: , R9, M, V, and w for the width of the loop. Do not substitute numerical values; use variables only) net = -mg (d) Wirite out Newton's second law for the loup. (Use the following as necessary: 4, R., mv, and w for the width of the loop. Do not substitute numerical values; use variables only.) (e) Obtain an expression for the speed of the loop as a function of time. (Use the following as necessary: B. R. ., v. t, and w for the width of the loop. Do not substitute numerical values; use variables only.) - - (f) Integrate the expression obtained in Part (e) to find the distance the loop falls as a function of time. (Use the following as necessary: 9, R, G, m, v, t, and w for the width of the loop. Do not substitute numerical values; use vanables anly.) Vịt) =

Answer 1

- solition: width of loop W=0.300m length 1 = 1.50m, B=0.2007 mass m = 50.og, Resistance R = 0.2002 Speed VIO) = 0., uro) = 0.100 m I OB Flun through the loop 0 = B (2-x)W 9 Induced EMF E= -2 E = BN du = Bwv Direction of induced EMF is clockwise. Induced current is also clockwise. I = e = BWV (b) FITXB Fonce on top edge of loop F = I WB (By right hand rule, f is upward) R Forces on other two edges get cancelled. fu= B2W v upward.

Fuet = mg-fu wa thet = mg- mg second law Newton's fret = n av de maten mg - Shop on B²W2 du t = g-B2W2; V16) MR v g-B²w2 MR 1 = t MR v B2 B2W 2 MR B²W2 MR V -B²W2 e mě t - = e ý se mat B22 MRg . v (t) = mě t l B²W2

da imgrc -B2WZ MR at =/ R? ganz single lien ) at 10) 0 e mrt mgr / Xlf = nrol r(t) = x10) + B2wa (t. -B2w2 MR - B?w2 mr , = (+) X t - n10+ mgr r me t Il-e ay] B2 2 / t. BBW 2U