ELECTROMAGNETIC INDUCTION
29
29.1.
29.2.
IDENTIFY: Altering the orientation of a coil relative to a magnetic field changes the magnetic flux through the coil. This change then induces an emf in the coil. SET UP: The flux through a coil of N turns
SOURCES OF MAGNETIC FIELD
28
28.1.
! ^ EXECUTE: (a) r = ( 0.500 m ) i , r = 0.500 m ! ! ^ v r = vr^ i = -vrk j ^
! IDENTIFY and SET UP: Use Eq.(28.2) to calculate B at each point. ! ! ! ! ! qv r 0 qv r ^ r ^ B= 0 = , since r = . 4 r 2 4 r 3
ELECTRIC CHARGE AND ELECTRIC FIELD
21
21.1.
(a) IDENTIFY and SET UP: Use the charge of one electron ( -1.602 10 -19 C) to find the number of electrons required to produce the net charge. EXECUTE: The number of excess electrons needed to produce n
MAGNETIC FIELD AND MAGNETIC FORCES
27
27.1.
! IDENTIFY and SET UP: Apply Eq.(27.2) to calculate F . Use the cross products of unit vectors from Section 1.10. ! ^ j EXECUTE: v = ( +4.19 104 m/s ) i + ( -3.85 104 m/s ) ^ ! ^ (a) B = (1.40 T ) i !
DIRECT-CURRENT CIRCUITS
26
26.1.
26.2.
26.3.
IDENTIFY: The newly-formed wire is a combination of series and parallel resistors. SET UP: Each of the three linear segments has resistance R/3. The circle is two R/6 resistors in parallel. EXECUTE: T
CURRENT, RESISTANCE, AND ELECTROMOTIVE FORCE
25
25.1.
25.2.
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 amou
CAPACITANCE AND DIELECTRICS
24
24.1.
24.2.
24.3.
Q Vab SET UP: 1 F = 10 -6 F EXECUTE: Q = CVab = (7.28 10 -6 F)(25.0 V) = 1.82 10 -4 C = 182 C EVALUATE: One plate has charge + Q and the other has charge -Q . Q PA and V = Ed . IDENTIFY and SE
ELECTRIC POTENTIAL
23
ra = 0.150 m rb = (0.250 m) 2 + (0.250 m) 2 rb = 0.3536 m
23.1.
IDENTIFY: Apply Eq.(23.2) to calculate the work. The electric potential energy of a pair of point charges is given by Eq.(23.9). SET UP: Let the initial position
GAUSS'S LAW
22
^ E = E cos dA, where is the angle between the normal to the sheet n and the
22.1.
(a) IDENTIFY and SET UP:
electric field E . EXECUTE: In this problem E and cos are constant over the surface so
E = E cos dA = E cos A = (1