Question

1. Asteady, uniform magnetic field of magnitude B exists in the horizontal, shaded region. This field is directed downward, as indicated. A rectangular loop of rigid, conductive wire, of length L, width W mass m, and resistance R, is initially at rest on a horizontal, frictionless surface, with its east end located at the edge of the field, as shown. CAardnae em >XXXXX XXXXXX XXXXXXXXXX XXX XXXXXXX XX XXXXXXX. side 1 КХXXXX XXXXXXХ W XXXXXXX ХXXXXXXXXXXXX K XX) XXXXXXX XXXXXXXXXXXXX>" side 2 Then you begin to push the loop so that it has a constant eastward acceleration until its east end is a distance d into the field (where d < L). From that moment (call it t = 0), you maintain the loop's motion at a steady velocity v eastward. The list of known values: B L, W, m, R, v, d, b, u. Find algebraic expressions for the following, expressing your results using only known values, and for full credit, you must show that the units of your solutions are also correct From the moment you begin to push until the moment defined as t = 0, how much work have you done on the loop? a. b. As you push the loop into the field, what is the direction of the force that side 2 of the loop exerts on side 1? Why doesn't the loop get deflected as a result of this force? Suppose that the field strength varies according to this time function: B(t)= Be, where b is a known nbe and =0 ar the moletomal ). Wha magn c. direction? Cou ps awill still be outside the field utside the felsume that the reversal occurs while i and its left end is still the loop's right end is still in the

Answer 1

Solution (a) Let the loop PQTS is pushed through a distance into the magnetic field Bo then the flux linked with the portion of loop in this position o B, Wx. The X PK: side 1 side 2 rate of change flux do B Wv but according to Faraday's law- dt do dx = BoW dt BO Wv, where E is the emf dt induced across the ends P and S. Because of the emf there flow a current 'i 'within the loop that Now the Westward force experienced by the loop is given byi W R R. Hence the work done against the magnetic force to move the loop through distance d is W (extena) = Fxd=Bo wvd (b). the current in the wires PQ and TS is same but opposite in direction. The equal and opposite magnetic force per unit length between the sidel and side 2 of the wire is given by 2 F .Hence, the force experience by side 2 of the loop - R2 2T W 2TW 2T H Bo vW The loop not get deflected because the driving force in 2T R2 in South direction is F East ward direction counter balance the deformation of the loop. (c). Please note in this question the exponential term is not clear. Though we can do as follows cause the flow of current in the loop in opposite .PQ in opposite direction F- Bi'W, due to this The change in magnetic field changes flux that direction, because of this current I the force on force the distance moved by the loop can be calculated. Let it be L' the area still within the field B is W.d-WL'. Thus the ratio of inside outside areas can be calculated C