35 1 Model Apply the Galilean transformation of velocity a In the laboratory frame S the speed of the proton is Solve v 1 41 10 m s 6 2 1 41 10 m s 6 2 2 0 10 m s 6 The angle the velocity vector makes with the positive y axis is 1 41 10 m s 1 41 10 m s cid 71 b In the rocket frame S we need to first determine the vector v cid 71 cid 71 cid 71 v V v 1 00 10 1 41 10 1 41 10 1 m s tan i i j 6 6 6 6 6 45 The speed of the proton is Equation 34 1 yields m s 0 41 10 1 41 10 m s 6 i 6 j v 0 41 10 m s 6 2 1 41 10 m s 6 2 1 47 10 m s 6 The angle the velocity vector makes with the positive y axis is tan 1 6 0 41 10 m s 1 41 10 m s 6 16 2 35 2 Model Apply the Galilean transformation of fields Visualize Please refer to Figure EX35 2 Solve a Equation 35 11 gives the Galilean field transformation equation for magnetic fields cid 71 B cid 71 cid 71 V E cid 71 B cid 71 B cid 71 V is in the positive k direction must be in the negative i direction so that cid 71 B Bk must be in the negative k direction Since VEk Ej The rocket scientist will measure B cid 71 E Ej 1 2 c cid 71 For B B V E cid 71 cid 71 V E Vi cid 71 cid 71 B if the rocket moves along the x axis cid 71 b For B B V E y axis or the y axis cid 71 c For B B V E moves along the x axis cid 71 must be zero The rocket scientist will measure B B if the rocket moves along either the must be in the positive k direction The rocket scientist will measure B B if the rocket 35 3 Model Use the Galilean transformation of fields Solve Equation 35 11 gives the Galilean transformation equations for the electric and magnetic fields in S and S frames cid 71 E cid 71 cid 71 cid 71 E V B cid 71 B cid 71 B cid 71 cid 71 V E 1 2 c In a region of space where cid 71 B 6 1 0 10 m s cid 71 0 i 1 2 c cid 71 cid 71 0B cid 71 E cid 71 E 6 j 1 0 10 V m The magnetic field is 6 1 0 10 V m j 1 0 10 12 8 3 0 10 2 k T 1 11 10 5 k T 35 4 Model Use the Galilean transformation of fields Visualize Please cid 71 E Solve Equation 35 11 gives the Galilean transformation equations for the electric and magnetic fields in S and S frames to Figure EX35 4 We are given 6 k 1 0 10 V m 2 0 10 m s 1 0 T j refer and cid 71 V 6 i B cid 71 cid 71 cid 71 E E V B cid 71 cid 71 cid 71 B B cid 71 cid 71 V E 1 2 c The electric and magnetic fields viewed from earth are cid 71 E 1 2 c 1 0 10 V m 2 0 10 m s k i 6 6 1 0 T j 6 1 0 10 V m k cid 71 B 1 0 T j 6 2 0 10 m s i 1 0 10 V m 1 0 T k j 6 12 2 0 10 V m 2 3 0 10 m s 8 j 0 99998 T j B B you need 5 significant figures of accuracy to tell the difference between them Assess Although 35 5 Model Use the Galilean transformation of fields Visualize Please cid 71 E 10 V m refer i j 6 to Figure EX35 5 We are given 1 2 1 2 cid 71 V 6 i 1 0 10 m s k 0 50 T and cid 71 B Solve Equation 35 11 gives the Galilean transformation equation for the electric field in the S and S frames cid 71 E cid 71 cid 71 cid 71 E V B cid 71 E The electric field from the moving rocket is 6 1 0 10 m s 0 707 10 V m i j 6 i 0 50 T k 0 707 10 i 6 0 207 10 V m 6 j tan 1 6 0 207 10 V m 0 707 10 V m 6 16 3 above the x axis 35 6 Model The net magnetic flux over a closed surface is zero Visualize Please refer to Figure EX35 6 Solve Because we can t enclose a net pole within a surface uniform over each face of the box the total magnetic flux around the box is m cid 71 cid 71 B dA 0 Since the magnetic field is 1 cm 2 cm 2 T 2 T 1 T 3 T cid 71 B unknown cid 71 dA unknown 0 0006 T m cid 71 The angle must be 180 Because is the angle between B into the face 1 cm 1 cm 2 T unknown 0 2 B unknown cos and the outward normal of d 6 T cid 71 cid 71 B is directed A the field 35 7 Solve The units of ed 0 dt are 2 C N m 2 2 N C m s C s A 35 8 Solve The displacement current is defined as Idisp 0 d e dt The electric flux inside a capacitor with plate area A is e EA The electric field inside a capacitor is E 0 Q A 0 and thus the electric flux is where VC is the capacitor voltage The capacitance C is constant hence the displacement current is e EA Q CV C 0 0 I disp 0 0 e d dt d CV C dt 0 C dV C dt 35 9 Model The displacement current is numerically equal to the current in the wires leading to and from the capacitor Solve During the charging process a parallel plate capacitor develops charge as a function of time If the charge on a capacitor plate is Q at time t then Q CVC where VC is the voltage across the capacitor plates Taking the derivative dQ dt I C dV C dt dV C dt I C 1 0 A 1 0 F 1 0 10 V s 6 35 10 Model The electric field inside a parallel plate capacitor is uniform As the capacitor is charged the changing electric field induces a magnetic field Visualize The induced magnetic field lines are circles concentric with the capacitor Please refer to Figure 35 18 Solve a The Ampere Maxwell law is …
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