exam3sol(1) - PHY2049 Fall 2010 Profs. S. Hershfield, A....

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Unformatted text preview: PHY2049 Fall 2010 Profs. S. Hershfield, A. Petkova Exam 3 Solution 1. The current i increases in the loop labeled in the figure. What is the direction of the induced current in the smaller inner loop and the loop at the right? (List the direction of the induced current for the smaller loop first.) Answer: clockwise, counterclockwise Solution: The magnetic field due to the current i is going out of the page inside the loop and into the page outside the loop. Since the current is increasing, the flux going out of the page increases for the small loop, and the flux into the page increases for the right loop. To oppose this (Lenzs law) the induced current for the small loop is clockwise and for the right loop is counterclockwise. 2. A square loop of side 2 cm moves with velocity 3 m/s at t = 0 as it exits a region of uniform magnetic field 5 T into the page (see figure). If the loop has resistance 0.4 , what is the magnitude of the magnetic force on it at t = 0? Answer: 7 . 5 10 2 N Solution: In this problem the flux through the square loop is decreasing because the area inside the loop where there is a magnetic field is decreasing. The rate of change of the area is dA/dt = (2 cm )(3 m/s ) = 0 . 06 m 2 /s , and the rate of change in flux is d B /dt = BdA/dt = 0 . 3 V . Consequently, the induced current is 0 . 3 V/ . 4 = 0 . 75 A . The induced current is clockwise to oppose the change in flux. In particular there is a current going upwards on the left side of the loop. Using vector F = i vector L vector B , the force on that segment of the loop is (0 . 75 A )(0 . 02 m )(5 T ) = 0 . 075 N to the left. The other segments create forces which cancel, and hence this is also the net force on the loop. 3. A generator consists of a circular coil of wire with 500 turns and radius 2 cm. It rotates in a uniform magnetic field of magnitude 3 T with frequency f = 60 Hz. What is the voltage or emf amplitude produced by the generator? Answer: 711 V Solution: In this problem the angle between the coil and the magnetic field is changing: B ( t ) = 500(3 T )( (0 . 02 m ) 2 ) cos( t ), where = 2 f is the angular frequency. Taking the derivative of the flux and using Faradays law the induced emf is- d B ( t ) /dt = (1 . 88 Wb ) sin( t ). The amplitude is (1 . 88 Wb ) = 710 V . 4. The magnetic field is 1 T into the page and 2 T out of the page in regions 1 and 2 respectively (see figure). The magnitude of both fields is increasing at a rate of 0.005 T/s . If region 1 has area 0.4 m 2 and region 2 has area 0.8 m 2 , what is contintegraltext vector E dvectors for the path of integration shown in the figure?for the path of integration shown in the figure?...
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exam3sol(1) - PHY2049 Fall 2010 Profs. S. Hershfield, A....

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