In Figure , two straight conducting rails form a right angle. A conducting bar in contact with the rails starts at the vertex at time t = 0 and moves with a constant velocity of 5.20 m/s along them. A magnetic field with B = 0.350 T is directed out of the page. Calculate (a) the flux through the triangle formed by the rails and bar at t = 3.00 s and(b) the emf around the triangle at that time. (c) If the emf is \\mathscr{E}=at^n , where a and n areconstants, what is the value of n?
(a) The "height" of the triangular area enclosed by the rails and bar is the same as the distance traveled in time v: d=v t, where v=5.20 \\mathrm{m} / \\mathrm{s} . We also note that the "base" of that triangle (the distance between the intersection points of the bar with the rails) is 2 d Thus, the area of the triangle is A=\\dfrac{1}{2}(\ ext { base})(\ ext { height })=\\dfrac{1}{2}(2 v t)(v t)=v^{2} t^{2} since the field is a uniform B=0.350\\mathrm{T} , then the magnitude of the flux (in SI units) is \\Phi_{B}=B A=(0.350)(5.20)^{2}t^{2}=9.46 t^{2} At t=3.00 \\mathrm{s}, we obtain \\Phi_{B}=85.2\\mathrm{Wb} (b) The magnitude of the emf is the(absolute value of) Faraday's law: \\varepsilon=\\dfrac{d \\Phi_{B}}{dt}=9.46 \\dfrac{d t^{2}}{d t}=18.9 t in SI units. At t=3.00 \\mathrm{s} ,this yields \\varepsilon=56.8 \\mathrm{V} (c) Our calculation in part (b) shows that n=1