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Unformatted text preview: Chapter 32: Maxwell’s 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 μ 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 μ Wb. Solution Gauss’ law for magnetic fields asserts that the net magnetic flux through a closed surface is zero. Therefore Φ ends + Φ curved surface = 0 or Φ curved surface =  Φ ends Since the magnetic field is uniform on each end the flux on each end can be calculated from Φ end = BA cos( θ ) where θ is 0 for the end where the field is directed outward and 180 ° for the end where the field is inward and A = π R 2 = 0.045 m 2 for each end. Therefore Φ curved surface =  (0.045 m 2 )*(1.60 mT) + 25.0 μ Wb =  47 μ 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 = μ o ε o R 2 2r dE dt 322 Maxwell’s Equations; Magnetism of Matter Therefore the rate of change of electric field between the plates is dE dt = 2rB μ o ε o R 2 dE dt = 2 * 6 mm ( ) * 2 × 107 T ( ) 4 π × 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 π (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 ∫ = μ o i enc + i d ( ) The integral on the left is equal to 2 π RB at the perimeter. On the right, i enc is zero and i d = ε o A dE dt = ε o A d V / x ( ) dt = ε o A x dV dt = ε o A x V max ∗ 120 π rad/s ( ) ∗ cos 2 π 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 π R μ ε o A x V max ∗ 120 π rad/s ( ) = μ ε o π R x V max ∗ 60 rad/s ( ) B max = μ ε o π 30 mm ( ) 5 mm 150 volt ( ) ∗ 60 rad/s ( ) = 1.9 pT Chapter 32: Maxwell’s Equations; Magnetism of Matter 323 (b) Apply AmpereMaxwell for r < R, 2 π rB max = μ ε o π r 2 x V max ∗ 120 π rad/s ( ) B max = μ ε o r 2 x V max ∗ 120 π rad/s ( ) Therefore if we plot B...
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This note was uploaded on 08/03/2009 for the course PHYSICS 2049 taught by Professor C during the Spring '09 term at Vanderbilt.
 Spring '09
 C
 Physics, Magnetism

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