Solve a initially and and the total energy stored is

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Solve: (a) Initially and and The total energy stored is (b) The current is maximum when and so and i max 5 Å 2 1 6.40 3 10 2 4 J 2 3.75 3 10 2 3 H 5 0.584 A. 1 2 Li max 2 5 6.40 3 10 2 4 J U L 5 6.40 3 10 2 4 J. U C 1 U L 5 6.40 3 10 2 4 J U C 5 0. q 5 0 0.640 mJ. U C 5 1 2 C v 2 5 1 2 1 5.00 3 10 2 6 F 21 16.0 V 2 2 5 6.40 3 10 2 4 J. U L 5 0 i 5 0. v 5 16.0 V U L 5 1 2 Li 2 . U C 5 1 2 C v 2 . v 5 1 " 1 5.25 3 10 2 3 H 21 12.0 3 10 2 6 F 2 5 3980 rad / s v 5 1 " LC . t 5 2 L ln 1 1 2 1 / " 2 2 R 5 2 1 1.50 3 10 2 3 H 2 ln 1 1 2 1 / " 2 2 0.750 3 10 3 V 5 2.46 m s. 1 " 2 5 1 2 e 2 1 R / L 2 t t 5 2 L ln 1 1 2 2 R 5 2 1 1.50 3 10 2 3 H 2 ln 1 1 2 2 0.750 3 10 3 V 5 1.39 m s. 2 1 R L 2 t 5 ln 1 1 2 2 . e 2 1 R / L 2 t 5 1 2 . 1 2 5 1 2 e 2 1 R / L 2 t i 5 1 2 E R E R . 1 ! 2 1 2 U 5 1 2 Li 2 . i 5 E R 1 1 2 e 2 1 R / L 2 t 2 . t 5 L R 5 0.65 ms. R 5 2 L ln 1 0.100 2 t 5 2 1 10.4 3 10 2 6 H 2 ln 1 0.100 2 1.50 3 10 2 3 s 5 1.60 3 10 2 2 V . 2 R L t 5 ln 1 0.100 2 0.100 5 e 2 1 R / L 2 t . 0.100 I 0 5 I 0 e 2 1 R / L 2 t i 5 0.100 I 0 . v 1 t 25.0 V v 2 t 25.0 V i t 1.67 A v 1 5 iR v 2 v 1 Electromagnetic Induction 21-13
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21.61. Set Up: Wires A and C have a length of 0.500 m and wire D has a length of where v is the component of velocity perpendicular to the field direction and perpendicular to the bar. Solve: (a) is parallel to so the induced emf is zero. (b) is perpendicular to The component of perpendicular to the bar is (c) is perpendicular to The component of perpendicular to the bar is Reflect: The induced emf depends on the angle between and and also on the angle between and the bar. 21.62. Set Up: The energy stored in a capacitor is The energy stored in an inductor is Energy conservation requires that the total stored energy be constant. The current is a maximum when the charge on the capacitor is zero. Solve: (a) At this instant (b) At this instant (c) This is when 21.63. Set Up: Use to find the amount of heat energy that must go into the water. The energy stored in an inductor is Solve: This much energy must be stored in the inductor, so This is a huge current; not reasonable for ordinary inductors. Reflect: Inductors are not useful devices for storing large amounts of electrical energy. Also, inductors typically are solenoids with many turns of wire and therefore have some resistance. If the current is large then the rate of loss of electrical energy will be very large. 21.64. Set Up: When the loop is entering or leaving the region of magnetic field the flux through it is changing and there is an induced emf. The magnitude of this induced emf is as in Example 21.4. The length L is 0.750 m. When the loop is totally within the field the flux through the loop is not changing and there is no induced emf. The induced current has magnitude and direction given by Lenz’s law. Solve: (a) The flux through the loop is directed out of the page and is increasing, so the magnetic field of the induced current is into the page inside the loop and the induced current is clockwise.
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