SOLUTIONS, ASSIGNMENT 26
29.28. IDENTIFY:Use Eq.(29.10) to calculate the induced electric field E at a distance r from the center of the solenoid. Away from the ends of the solenoid, B = 0 nI inside and B = 0 outside. (a) SET UP:The end view of the soleno
SOLUTIONS, ASSIGNMENT 25
29.15. IDENTIFY and SET UP:The field of the induced current is directed to oppose the change in flux. EXECUTE:(a) The field is into the page and is increasing so the flux is increasing. The field of the induced current is out of t
Solutions, Assignment 24
29.6.IDENTIFY:Apply Eq.(29.4). I = E/R. SET UP: d B /dt = AdB/dt. EXECUTE:(a) E = Nd B d d = NA ( B ) = NA ( (0.012 T/s)t + (3.00 105 T/s 4 )t 4 ) dt dt dt
E = NA ( (0.012 T/s) + (1.2 104 T/s 4 )t 3 ) = 0.0302 V + (3.02 104 V/s3 )
SOLUTIONS, Assignment 23
28.40. = IDENTIFY: B nI0 = 0 NI L
SET UP: L = 0.150 m 0 (600) (8.00 A) = 0.0402 T EXECUTE: B = (0.150 m) EVALUATE:The field near the center of the solenoid is independent of the radius of the solenoid, as long as the radius is muc
SOLUTIONS, ASSIGNMENT 22
28.66. r r 0 Idl r IDENTIFY:Apply dB = . 4 r 2 SET UP:The two straight segments produce zero field at P. The field at the center of a circular loop of I I radius R is B = 0 , so the field at the center of curvature of a semicircul
Solutions, Assignment 21
28.11.
IDENTIFY and SET UP:The magnetic field produced by an infinitesimal current element is given by Eq. (28.6). r r 0 Il r As in Example 28.2 use this equation for the finite 0.500-mm segment of wire since the dB = 2 4 r l = 0.
Solutions, Assignment 20
27.42. IDENTIFY: = IAB sin . The magnetic moment of the loop is = IA . SET UP:Since the plane of the loop is parallel to the field, the field is perpendicular to the normal to the loop and = 90 . EXECUTE:(a) = IAB = (6.2 A)(0.050
SOLUTIONS, ASSIGNMENT 19
27.28. IDENTIFY:For no deflection the magnetic and electric forces must be equal in magnitude and opposite in direction. 2 SET UP: v = E / B for no deflection. With only the magnetic force, q vB = mv / R EXECUTE:(a) v = E B = (1.5
SOLUTIONS, ASSIGNMENT 18
r 27.1.IDENTIFY and SET UP:Apply Eq.(27.2) to calculate F . Use the cross products of unit vectors from Section 1.10. r 4 4 j EXECUTE: v = ( +4.19 10 m/s ) i + ( 3.85 10 m/s ) r (a) B = ( 1.40 T ) i r r r F = qv B = ( 1.24 108 C )
SOLUTIONS, ASSIGNMENT 17
26.38. IDENTIFY:An uncharged capacitor is placed into a circuit. Apply the loop rule at each time. SET UP:The voltage across a capacitor is VC = q / C . EXECUTE:(a) At the instant the circuit is completed, there is no voltage over
SOLUTIONS, ASSIGNMENT 16 26.2. IDENTIFY:It may appear that the meter measures X directly. But note that X is in parallel with three other resistors, so the meter measures the equivalent parallel resistance between ab. SET UP:We use the formula for resisto
Solutions, Assignment 15
25.33. IDENTIFY: V = E Ir . SET UP:The graph gives V = 9.0 V when I = 0 and I = 2.0 A when V = 0. EXECUTE:(a) E is equal to the terminal voltage when the current is zero. From the graph, this is 9.0 V. (b) When the terminal voltag
SOLUTIONS, ASSIGNMENT 14
25.1. IDENTIFY: I = Q / t . SET UP: 1.0 h = 3600 s EXECUTE: Q = It = (3.6 A)(3.0)(3600 s) = 3.89 104 C. EVALUATE:Compared to typical charges of objects in electrostatics, this is a huge amount of charge. (a) IDENTIFY:By definition
SOLUTIONS, ASSIGNMENT 13
24.30.
0 A Q2 CV 2 . The stored energy can be expressed either as or as , whichever is more d 2C 2 convenient for the calculation. SET UP:Since d is halved, C doubles. EXECUTE:(a) If the separation distance is halved while the cha
SOLUTIONS, ASSIGNMENT 12
24.10. IDENTIFY:Capacitance depends on the geometry of the object. (a) SET UP:The capacitance of a cylindrical capacitor is C = rb = ra e 2 P0 L / C . EXECUTE:Substituting in the numbers for the exponent gives = 0.182 3.67 1011 F
Solutions, Assignment 11
24.1.IDENTIFY: C =
Q Vab
SET UP: 1 F =106 F EXECUTE: Q = CVab = (7.28 106 F)(25.0 V) = 1.82 104 C = 182 C EVALUATE:One plate has charge +Q and the other has charge Q . 24.2. IDENTIFY and SET UP: C = (a) C = P 0
PA Q 0 , C = and V
SOLUTIONS, ASSIGNMENT 10
kq . The total potential at any point is the algebraic sum of the r
23.25.
IDENTIFY:For a point charge, V =
potentials of the two charges. SET UP:(a) The positions of the two charges are shown in Figure 23.25a.
Figure 23.25a (b) x
SOLUTIONS, ASSIGNMENT 9 plus Problem 23.25
23.10. IDENTIFY:The work done on the alpha particle is equal to the difference in its potential energy when it is moved from the midpoint of the square to the midpoint of one of the sides. SET UP:We apply the for
SOLUTIONS, ASSIGNMENT 8
22.51. IDENTIFY:The net electric field is the vector sum of the fields due to the sheet of charge on each surface of the plate. SET UP:The electric field due to the sheet of charge on each surface is E = / 2P and is directed 0 away
SOLUTIONS, ASSIGNMENT 7
22.13. rr E (a) IDENTIFY and SET UP:It is rather difficult to calculate the flux directly from = dA since the r r magnitude of E and its angle with dA varies over the surface of the cube. A much easier approach is to use Gauss's la
SOLUTIONS, ASSIGNMENT 6
22.3. IDENTIFY:The electric flux through an area is defined as the product of the component of the electric field perpendicular to the area times the area. (a) SET UP:In this case, the electric field is perpendicular to the surface
SOLUTIONS, ASSIGNMENT 5
21.55. IDENTIFY:For a ring of charge, the electric field is given by Eq. (21.8). F = qE . In part (b) use Newton's third law to relate the force on the ring to the force exerted by the ring. SET UP: Q = 0.125 109 C, a = 0.025 m and
ASSIGNMENT 4
k q1q2 and solve for r. r2 SET UP: F = 650 N . EXECUTE: r = k q1q2 (8.99 109 N m 2 /C2 )(1.0 C)2 = = 3.7 103 m = 3.7 km F 650 N
21.5.IDENTIFY:Apply F =
EVALUATE:Charged objects typically have net charges much less than 1 C. 21.44. IDENTIFY:Fo
ASSIGNMENT 3
21.4.IDENTIFY:Use the mass m of the ring and the atomic mass M of gold to calculate the number of gold atoms. Each atom has 79 protons and an equal number of electrons. SET UP: N A = 6.02 1023 atoms/mol . A proton has charge +e. EXECUTE:The m
+Assignment 2 solutions 21.2.IDENTIFY:The charge that flows is the rate of charge flow times the duration of the time interval. SET UP:The charge of one electron has magnitude e = 1.60 10-19 C. EXECUTE:The rate of charge flow is 20,000 C/s and t = 100 s =