Solve 2129 set up and where is the flux through one

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Solve: 21.29. Set Up: and where is the flux through one turn of the second coil. Solve: (a) (b) (c) Reflect: We can express M either in terms of the total flux through one coil produced by a current in the other coil, or in terms of the emf induced in one coil by a changing current in the other coil. E 1 5 M P D i 2 D t P 5 1 6.82 3 10 2 3 H 21 0.360 A / s 2 5 2.46 3 10 2 3 V 5 2.46 mV F B 2 5 Mi 1 N 2 5 1 6.82 3 10 2 3 H 21 1.20 A 2 25 5 3.27 3 10 2 4 Wb M 5 E 2 0 D i 1 / D t 0 5 1.65 3 10 2 3 V 0.242 A / s 5 6.82 3 10 2 3 H 5 6.82 mH F B 2 M 5 P N 2 F B 2 i 1 P , E 2 5 M P D i 1 D t P . E 1 5 M P D i 2 D t P M 5 1 4 p 3 10 2 7 Wb / m # A 21 800 21 50 2 p 1 0.200 3 10 2 2 m 2 2 0.100 m 5 6.32 3 10 2 6 H 5 6.32 m H M 5 m 0 N 1 N 2 A l M 5 m 0 N 1 N 2 A 2 p r 5 1 4 p 3 10 2 7 Wb / m # A 21 100 21 500 21 4.00 3 10 2 4 m 2 2 2 p 1 0.100 m 2 5 4.00 3 10 2 5 H. F B 2 5 B 1 A 5 m 0 N 1 i 1 A 2 p r . B 1 5 m 0 N 1 i 1 2 p r . M 5 P N 2 F B 2 i 1 P . R 5 p d 2 4 1 E Bd 2 5 p d E 4 B . v 5 E Bd volume / time 5 R 5 p d 2 v / 4. p 1 d 2 / 4 2 v t . A 5 p d 2 / 4. B 5 E v d 5 1.0 3 10 2 3 V 1 0.15 m / s 21 5.0 3 10 2 3 m 2 5 1.3 T E 5 v Bd . E 5 v BL F ext 5 0.800 N F S I F I 5 ILB sin f 5 1 2.00 A 21 0.500 m 21 0.800 T 2 sin 90° 5 0.800 N. I 5 E R 5 3.00 V 1.50 V 5 2.00 A. B S E 5 v BL 5 1 7.50 m / s 21 0.800 T 21 0.500 m 2 5 3.00 V F I F ext E 5 v BL . 21-8 Chapter 21
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21.30. Set Up: The emf in solenoid 2 produced by changing current in solenoid 1 is given by The mutual inductance of two such solenoids is derived in Example 21.8 to be where A and l are the cross-sectional area and length of solenoid 1. Solve: (a) (b) 21.31. Set Up: Solve: (a) is constant, so is constant. (b) with the same M as in part (a), so 21.32. Set Up: The inductance L is related to the flux through one turn by The voltage across the coil is related to the rate at which the current in it is changing by Solve: (a) (b) 21.33. Set Up: Solve: 21.34. Set Up: Solve: 21.35. Set Up: Example 21.10 shows that the inductance of a toroidal solenoid is The voltage across the coil is related to the rate at which the current in it is changing by Solve: (a) (b) Reflect: The inductance is determined solely by how the coil is constructed. The induced emf depends on the rate at which the current through the coil is changing. 21.36. Set Up: Example 21.10 shows that the inductance of a toroidal solenoid is Solve: L is proportional to so increasing N to doubles L. 21.37. Set Up: The magnetic field inside a long solenoid is uniform and equal to Solve: Reflect: The self-inductance depends only on the geometrical parameters of the coil and is independent of the cur- rent that flows in the coil. L 5 N F B i 5 m 0 N 2 A l . F B 5 BA . B 5 m 0 N l i . L 5 P N F B i P . " 2 N N 2 , L 5 m 0 N 2 A 2 p r . P D i D t P 5 E L 5 2.00 V 2.50 3 10 2 3 H 5 800 A / s N 5 Å 2 p rL m 0 A 5 Å 2 p 1 0.0600 m 21 2.50 3 10 2 3 H 2 1 4 p 3 10 2 7 T # m / A 21 2.00 3 10 2 4 m 2 2 5 1940 turns E 5 L P D i D t P . L 5 m 0 N 2 A 2 p r . E 5 1 0.260 H 21 18.0 3 10 2 3 A / s 2 5 4.68 3 10 2 3 V 5 4.68 mV E 5 L P D i D t P L 5 E 0 D i / D t 0 5 0.0160 V 0.0640 A / s 5 0.250 H E 5 L P D i D t P P D i D t P 5 E L 5 1.16 V 4.5 3 10 2 3 H 5 258 A / s F B 5 Li N 5 1 4.5 3 10 2 3 H 21 11.5 A 2 125 5 4.1 3 10 2 4 Wb E 5 L P D i D t P .
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