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1. Current Continuity and Relaxation Time P4.4: At t = 0 seconds, 60.0 C is evenly distributed throughout a 2.00 cm diameter pure silicon sphere. (a) Find the initial charge density. (b) How long does it take to drop to 10% of its initial value? (c) What will be the final surface charge density? 4 Q C 3 (a) Volume v 0.01m 4.19 x106 m 3 ; vo 14.3 3 3 v m t / (b) v 0.10 vo vo e 11.8 8.854 x10 For pure silicon, 4.4 x104 ln 0.10 t 12 237.4 x10 9 sec 2.303; t 2.303 547ns Fig. P4.8 (at end of animation) Q 60C ; area=4 r 2 1.257 x103 m 2 ; s 47.8 mC 2 (c) s m area 1.257 x103 m 2 3. Faraday’s Law and Transformer EMF P4.9: The magnetic flux density increases at the rate of 10 (Wb/m 2)/sec in the z direction. A 10 cm x 10 cm square conducting loop, centered at the origin in the x-y plane, has 10 ohms of distributed resistance. Determine the direction (with a sketch) and magnitude of the induced current in the conducting loop. dB Wb 10 2 a z dt ms dB Wb Vemf dS 10 2 a z dxdya z dt ms Wb Vs Vemf 0.1 0.1V s Wb 0.1V 10mA I= I 10 Fig. P4.9 I=10mA clockwise (when viewed From +z) P4.13: A 1.0 mm diameter copper wire is shaped into a square loop of side 4.0 cm. It is placed in a plane normal to a magnetic field increasing with time as B = 1.0 t Wb/m2 az, where t is in seconds. (a) Find the magnitude of the induced current and indicate its direction in a sketch. (b) Calculate the magnetic flux density at the center of the loop resulting from the induced current, and compare this with the original magnetic flux density that generated the induced current at t = 1.0 sec. We find the distributed resistance of the loop and work the problem assuming this resistance is lumped in one spot as shown in the figure. (a) The induced current is Vemf divided by the distributed resistance of the wire loop. 4 0.04m) 1 l 1m R 3.5m 7 A 5.8 x10 0.0005m 2 dB Wb dB Wb 1 2 a z ; Vemf dS 1 2 dt ms dt ms I ind 0.04 0.04 0 0 dx dy 1.6mV 1.6mV 0.46 A (note that this answer has no time dependence) 3.5m Fig. P4.13 (b) The field at the center of the loop from a single arm of the loop is found from Eqn. (3.7): a I z I H 2 2 a z 2 4 z 2 a So B 4o H 13 Wb m2 0.46 1 1 A -a z a z 2.59a z ; 2 m 2a 2 0.02 az . P4.15: A triangular wire loop has its vertices at the points (2, 0, 0), (0, 3, 0) and (0, 0, 4), with dimensions in meters. A time-varying magnetic field is given by B = 4t ay Wb/m2 (t in seconds). If the wire has a total distributed resistance of 2 , calculate the induced current and indicate its direction in a carefully drawn sketch. B dS, t B Wb 4a y 2 t ms 1 MN S MN 2 MN Vemf M 2a x 3a y , N 2a x 4a z S 6a x 4a y 3a z m2 Vemf = -(4)(4)=-16V Vemf 16V I 8A R 2 Fig. P4.15 4. Faraday’s Law and Motional EMF P4.22: Consider the rotating conductor shown in Figure 4.24. The center of the 2a diameter bar is fixed at the origin, and can rotate in x-y plane with B = Boaz. The outer ends of the bar make conductive contact with a ring to make one end of the electrical contact to R; the other contact is made to the center of the bar. Given Bo = 100. mWb/m2, a = 6.0 cm, and R = 50. , determine I if the bar rotates at 1.0 revolution per second. d a a , dL d a dt Figure P4.22 indicates one of the paths for the circulation integral. a a Bo a 2 Vemf a Boa z d a Bo d 2 0 0 Vemf I u B dL; Vemf u Bo a 2 R I 22.6 A 1 rev rad 2 1 Vs A 3 Wb 1 0.06m 2 100 x10 2 2R 2 s rev m 50 Wb V Fig. P4.22 P4.24: Consider a sliding rail problem where the conductive rails expand as they progress in the y direction as shown in Figure 4.25. If w = 10. cm and the distance between the rails increases at the rate of 1.0 cm in the x direction per 1.0 cm in the y direction, and uy = 2.0 m/sec, find the Vemf across a 100. resistor at the instant when y = 10. cm if the field is Bo = 100. mT. First we modify the figure so that the top rail is horizontal and all the spreading occurs via the bottom rail. As before, our approach will be to find and then d /dt. We have: B dS Boa z dxdya z Now, notice that x and y are not independent and are in fact related: x=y+w So we have y yw y 1 Bo dxdy Bo y w dy Bo y 2 wy 2 y 0 x 0 0 d dy Wb m Bo y w Bo y w u y 0.100 2 0.1m 0.1m 2 dt dt m s 40mV Vemf Vemf Alternate Method: Vemf u B dL u y a y Boa z dxa x Vemf u y Bo 1 y 2 dx u y Bo w y 1 w y 2 Fig. P4.24 5. Displacement Current P4.29: A 1.0 m long coaxial cable of inner conductor diameter 2.0 mm and outer conductor diameter 6.0 mm is filled with an ideal dielectric with r = 10.2. A voltage v(t) = 10.cos(6x106 t) mV is placed on the inner conductor and the outer conductor is grounded. Neglecting fringing fields at the ends of the coax, find the displacement current between the inner and outer conductors. C Q , so Q Cv(t ) V for coax (from chapter 2): C= D dS D a 2 lVo cos t 2 l so Q ln b a ln b a d dza 2 l D Q 2 l Vo cos t Vo cos t V cos t and D o a . 2 l ln b a ln b a ln b a D Vo sin t a Jd t ln b a so D 2 l Vo sin t 1 2 l Vo sin t id J d dS d dz ln b a 0 ln b a 0 Now evaluate id with the given parameters: 2 6 x106 rad s 10.2 8.854 x1012 F m 1m 0.010V sin t C As id ln 0.006 0.002 FV C id 97 sin 6 x106 t A