Question

X X X X X X x x X x X X XL X x X X X x X Be paper x Two parallel rails are connected together at one end by a resistance of R - 15.00 as shown in the diagram to the right. Across these two rails, which are L - 35.0 cm apart, there lies a conducting metal bar. The magnetic field is uniform, has a strength of B - 2.50 Tesla and is directed into the paper as shown. A force is applied to the metal bar so as to push the bar to the right with a velocity v = of 6,40 m/s. 29. What will be the direction of the resulting conventional current flowing through this circuit, and in what direction will the induced magnetic field point due to the induced current (Two answers in the order in which they were asked) 30. What will be the resulting EMF in this circuit?

Answer 1

Therefore circuit in anticlorack wise direction as shown in х x x x x х X X X Y > V RE X x x х X L x x х x x cuorrenta Х х X х X x rod moves 29 assume the ad Borgo is Hence B 2 at we along x-anis so, 82 (6.40 m2s) and z anis nato pand out perpendiculor to the plane of page and points outward (2.509 T) (-Ã) ?(-2.50% ) know the corrent flows through the circuit due to elect flow of electron and & the direction of current is opposite to flow of electron. electron with the soor metal Now, We an - е ( x B') 2 Force on bor is e {(6.40 misj î x (-2.50 ){} 6 1.6x1019 X Ġ: 40 X 2.50 NJ(ixã 2 (-2-56 N) electron mover along negative y anis. the current flows through the + So A y-anis. Hence figure.

direction of According to Lena's law the induced magnetic field due to induced worent will be in that direction which these resists the induced current. Here to resistance the induced current the direction of magn induced magnetic field must be point out ward to the plane of paper. 30 The resulti ng conf EMF in this circuit & BL 16.40m/s) x (2:507) x(35 'Ox 1032m) 5.60V