Question

A rectangular loop of wire with dimensions and w is released from rest at time t = 0 from a region with zero magnetic field into a region with uniform magnetic field Bo pointing into the page. At t=0, the upper edge of the loop is in the zero field region and the lower edge is in the uniform field region as shown in the figure below. The loop has mass m and the acceleration due to gravity is g and points downwards. For all parts below, consider only times 0<t<T, where T is the time when the upper edge of the loop enters the uniform field region. B=0 W x x x x x x X X X X X X 18 xxxxxx B=B. For parts (a) to (c) assume that the self-inductance of the loop L = 0, and that the resistance of the loop R+0. (a) Write down an expression for the force on the wire at time. [1pt] (b) Using your answer to part (a), Newton's second law, and the initial conditions given in the problem, find the velocity of the loop as a function of time.* [1pt] (c) Using your answer to part (b), find the current in the loop as a function of time. [1pt] For parts (d) and (e) assume that R=0 but L+0. (d) Write down a second order linear differential equation for the velocity of the loop. [1 pt) (e) By solving the equation you found in part (d) using the initial conditions given in the problem, find both the current and velocity of the loop as a function of time. [1pt] * Recall that the solution to a linear differential equation with constant coefficients is the sum of the general solution to the homogeneous equation and a particular solution to the inhomogeneous equation.

Answer 1

X x x X B-Bo So, wrrent in the loop let velocity of loop at time. t is given by v=v(A) So em generated (Induced) In the loop is E = veB = u(t)e Bo it in the loop as a function of time I(t) = &- Bov (A) so the magnetic Force on wire. Em = I (H) x 2 x B = B V (+) Net Force on wire Fnet - Fg - im as R = = mg - B = Fret From Newton's Freet = ma =) 2nd law ma 9 – 12 B2 ma 2 mg - tror mange m dr dN are the et dt vet) dt ar MR — vo t MR ma 2 Vie) = my head (2 - exp(-203 t/mr)] 1 MR

so the loop as a the current in Function of time is I (t) = Bolv (4) Lp R 164)= [- exp(- let) 7 I Bo dt PT Rio but to than em generated (induced) in the circit is equilized by &- Bol v = – L dI i wrrent in the V(t) = -2 dl circuit Bol dt V(t) = V velocity of loop Net Force equation at time t on loop using Newton's 2nd lan Fnet = Fg - fom = ma IL inductance of loop dt a = acceleration Ilt) eBo ML d²I Bol dt2 = m d v -) m q I (A) e Bo - ma N dt her mg - Ilc. =0