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Unformatted text preview: Chapter 32: Maxwells Equations; Magnetism of Matter Problem 3 A Gaussian surface in the shape of a rightcircular cylinder with end caps has a radius of 12.0 cm and a length of 80 cm. Through one end there is an inward magnetic flux of 25.0 Wb. At the other end there is a uniform magnetic field of 1.60 mT, normal to the surface and directed outward. What is the net magnetic flux through the curved surface? Answer 47.0 Wb. Solution Gauss law for magnetic fields asserts that the net magnetic flux through a closed surface is zero. Therefore ends + curved surface = 0 or curved surface =  ends Since the magnetic field is uniform on each end the flux on each end can be calculated from end = BA cos( ) where is 0 for the end where the field is directed outward and 180 for the end where the field is inward and A = R 2 = 0.045 m 2 for each end. Therefore curved surface =  (0.045 m 2 )*(1.60 mT) + 25.0 Wb =  47 Wb Problem 5 The induced magnetic field 6.0 mm from the central axis of a circular parallelplate capacitor and between the plates is 2.0 107 T. The plates have radius 3.00 mm. At what rate d E /dt is the electric field between the plates changing? Answer 2.4 10 13 V/m/s Solution Since 6.00 mm is greater than the radius of the plates the magnetic field is given by B = o o R 2 2r dE dt 322 Maxwells Equations; Magnetism of Matter Therefore the rate of change of electric field between the plates is dE dt = 2rB o o R 2 dE dt = 2 * 6 mm ( ) * 2 107 T ( ) 4 107 H/m ( ) * 8.85 1012 F/m ( ) * 3 mm ( ) 2 dE/dt = 2.4 10 13 V/m/s Problem 7 Suppose that a parallelplate capacitor has circular plates with radius R = 30 mm and a plate separation of 5.0 mm. Suppose also that a sinusoidal potential difference with a maximum value of 150 V and a frequency of 60 Hz is applied across the plates; that is, V = (150 V) sin[2 (60 Hz)t]. (a) Find B max (R), the maximum value of the induced magnetic field that occurs at r = R. (b) Plot B max (r) for 0 < r < 10 cm. Answer (a) 1.9 pT. Solution (a) We can use the AmpereMaxwell law to find the magnetic field that appears at the perimeter of the gap between the two plates. B d s = o i enc + i d ( ) The integral on the left is equal to 2 RB at the perimeter. On the right, i enc is zero and i d = o A dE dt = o A d V / x ( ) dt = o A x dV dt = o A x V max 120 rad/s ( ) cos 2 60 Hz ( ) t ( ) The maximum value of B occurs when i d is a maximum and i c is a maximum when the cosine term is equal to 1. Therefore maximum B is B max = 1 2 R o A x V max 120 rad/s ( ) = o R x V max 60 rad/s ( ) B max = o 30 mm ( ) 5 mm 150 volt ( ) 60 rad/s ( ) = 1.9 pT Chapter 32: Maxwells Equations; Magnetism of Matter 323 (b) Apply AmpereMaxwell for r < R, 2 rB max = o r 2 x V max 120 rad/s ( ) B max = o r 2 x V max 120 rad/s ( ) Therefore if we plot B...
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 Spring '09
 C
 Physics, Magnetism

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