A rectangular circuit s moved at a constant velocity of 3.0m/s into, through, an
ID: 2197784 • Letter: A
Question
A rectangular circuit s moved at a constant velocity of 3.0m/s into, through, and then out of a uniform 1.25-T magnetic field. The magnetic-firld region is considerably wider than 50.0cm. Find the magnitude and direction(clockwise or counterclockwise) of the current induced in the circuit as it is a. going into the magnetic field. b.totally within the magnetic field but still moving c.moving out of the field. d.sketch a graph of the current in the circuit as a function of time, including the preceding three cases.Explanation / Answer
First lets find out the direction of the induced current. The magnetic field is upwards. As the radius of the loop is increasing, and so is its area, the magnetic flux in upward direction is increasing. To counter this, a current will be induced so as to oppose this flux. So the direction of the current will be such that the magnetic field is downwards, hence, the current will be clockwise when viewed from above. Since we know the radius as a function of time, we need the area of the loop as a function of time. A = pr² A = p(r? + vt)² By Faraday's law, E = d?/dt E = d(AB) / dt = B (dA / dt) (because B is constant) dA / dt = d(pr²) / dt = 2pr (dr/dt) = 2p(r? + vt)(v) (because dr/dt = v) At t = 4.69 dA / dt = 0.057 m²/sec E = 1.95 x 0.057 = 0.11 V I = V/R = 0.11 / 11.7 = 9.5 mA (Clockwise)