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solutions_chapter21

The magnetic force on a positive charge in side 2 is

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a positive charge in side 4 is toward the top of the page. The magnetic force on a positive charge in side 2 is toward the bottom of the page. The force in side 4 is larger since the magneitc field of the wire is larger there. The net force is directed so as to produce a clockwise current. (c) (i) When as it should. (ii) When When the area of the loop goes to zero and the flux through the loop goes to zero. 21.67. Set Up: Let the flip take time The charge that flows is Flipping the coil by changes by and Solve: Reflect: The total charge that flows is independent of the time D t . N 5 QR 2 BA 5 1 0.400 3 10 2 3 C 21 50.0 V 2 2 1 0.100 T 2 p 1 2.00 3 10 2 2 m 2 2 5 80. Q 5 I D t 5 2 BAN R . I 5 E R 5 2 BAN R D t . E 5 N P DF B D t P 5 2 BAN D t . D 1 cos f 2 5 2. 180° f 180° Q 5 I D t . D t . a S 0 E S 0. a S 0, E S 0, v S 0, E 5 E 1 2 E 2 5 m 0 I v b 2 p 1 1 r 2 1 r 1 a 2 5 m 0 I v ba 2 p r 1 r 1 a 2 . E 1 5 B 1 v b 5 m 0 I 2 p r v b E 2 5 B 2 v b 5 m 0 I 1 2 p 21 r 1 a 2 v b 1 3 4 2 b I r a v B S . v S v S E 5 0 v S E 5 v BL B 5 m 0 I 2 p r 21-16 Chapter 21
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21.68. Set Up: When the switch is closed current flows counterclockwise in the circuit on the left, from the positive plate of the capacitor. The current decreases as a function of time, as the charge and voltage of the capacitor decrease. The direction of the current induced in circuit A is given by Lenz’s law. Solve: At loop A the magnetic field from the wire of the other circuit adjacent to A is into the page. The magnetic field of this current is decreasing, as the current decreases. Therefore, the magnetic field of the induced current in A is directed into the page inside A and to produce a magnetic field in this direction the induced current is clockwise. 21.69. Set Up: The energy densities to each field are and Solve: says 21.70. Set Up: Closing and simultaneously opening produces an L - C circuit with initial current through the inductor of 3.50 A. When the current is a maximum the charge q on the capacitor is zero and when the charge q is a maximum the current is zero. Conservation of energy says that the maximum energy stored in the inductor equals the maximum energy stored in the capacitor. Solve: (a) (b) When is maximum, 21.71. Set Up: Inside the solenoid, Outside the solenoid, The energy density in the magnetic field is The total energy is Solve: (a) (b) The volume occupied by the magnetic field is Reflect: We could also calculate U from 21.72. Set Up: The emf in the inductor is is limited by the loop rule, so the rate of change of the current is limited and the current in the inductor can’t jump suddenly from one value to another. Solve: (a) The current in the inductor is zero when the switch is open and is still zero immediately after the switch is closed. The current through the resistor equals the current through the inductor and therefore is zero. The volt- age across the resistor must equal the battery voltage, so the current through this resistor is (b) After a long time the current is no longer changing and the voltage across the inductor is zero. The voltage across each resistor is 60.0 V, so the current through the resistor is 3.00 A and the current through the resistor is 2.00 A.
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