R = ρ L [ π( ) 2 = Ω. πr 2 = (10 A)( Ω) π(0.05 m) 2 =1.4 T/s.

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1 5. (a) Table 27-1 gives the resistivity of copper. Thus, R = ρ L [ ] A =(1.68 π(0.10 m) 10 8 Ω m) π( ) 2 = Ω. /4 (b) We use i = E /R = dφ B /dt /R =(πr 2 /R) db/dt. Thus db dt = ir πr 2 = (10 A)( Ω) π(0.05 m) 2 =1.4 T/s.

2 11. (a) Ohm s law combines with Faraday s law to give i = N dφ B R dt where R is the resistance of the coil. In this case, N = 1 (it is a single loop), and we integrate to find the charge: t 0 t idt = 1 dφ B dt R 0 dt q(t) = 1 R (Φ B(t) Φ B (0)) which is equivalent to the expression shown in the problem statement. We have used little more than the fundamental theorem of calculus; no particular assumptions have been made about how the integrations should be performed. The result is independent of the way B has changed. (b) If the current is identically zero for over the whole range 0 t then certainly the left-hand side of our computation, above, gives zero. But the same result can come from the current being in one directionfor,say,0 t 2 and then in the opposite direction for t 2 t in such a way that t 0 idt=0. So a vanishing integral does not necessarily mean the integrand itself is identically zero.

3 17. (a) It should be emphasized that the result, given in terms of sin(2πft), could as easily be given in terms of cos(2πft) orevencos(2πft + φ) whereφ is a phase constant as discussed in Chapter 16. The angular position θ of the rotating coil is measured from some reference line (or plane), and which line one chooses will affect whether the magnetic flux should be written as BAcos θ, BAsin θ or BAcos(θ + φ). Here our choice is such that Φ B = BAcos θ. Since the coil is rotating steadily, θ increases linearly with time. Thus, θ = ωt (equivalent to θ =2πft) ifθ is understood to be in radians (and ω would be the angular velocity). Since the area of the rectangular coil is A = ab, Faraday slawleadsto d(bacos θ) E = N = NBA d cos(2πft) = NBab2πf sin(2πft) dt dt which is the desired result, shown in the problem statement. The second way this is written (E 0 sin(2πft)) is meant to emphasize that the voltage output is sinusoidal (in its time dependence) and has an amplitude of E 0 =2πfNabB. (b) We solve E 0 = 150 V = 2πfNabB when f =60.0 rev/s and B =0.500 T. The three unknowns are N,a, and b which occur in a product; thus, we obtain Nab =0.796 m 2. This means, for instance, that if we wanted the coil to have a square shape and consist of 50 turns, then the side length of thesquarewouldbea = b =0.126 m.

4 35. We use Faraday s law intheform E d s = (dφb /dt), integrating along the dotted path shown in the Figure. At all points on the upper and lower sides the electric field is either perpendicular to the side or else it vanishes. We assume it vanishes at all points on the right side (outside the capacitor). On the left side it is parallel to the side and has constant magnitude. Thus, direct integration yields E d s = EL, where L is the length of the left side of the rectangle. The magnetic field is zero and remains zero, so dφ B /dt = 0. Faraday s law leads to a contradiction: EL = 0, but neither E nor L is zero. Therefore, there must be an electric field along the right side of the rectangle.

5 39. We refer to the (very large) wire length as l and seek to compute the flux per meter: Φ B /l. Using the right-hand rule discussed in Chapter 30, we see that the net field in the region between the axes of antiparallel currents is the addition of the magnitudes of their individual fields, as given by Eq and Eq There is an evident reflection symmetry in the problem, where the plane of symmetry is midway between the two wires (at x = d/2); the net field at any point 0 <x<d/2isthesameat its mirror image point d x. The central axis of one of the wires passes through the origin, and that of the other passes through x = d. We make use of the symmetry by integrating over 0 <x<d/2and then multiplying by 2: Φ B =2 d/2 0 BdA=2 a 0 B (ldx)+2 d/2 a B (ldx) where d = m is diameter of each wire. We will r instead of x in the following steps. Thus, using the equations from Ch. 30 referred to above, we find Φ B l = 2 a 0 = µ 0i 2π ( ) µ0 i 2πa 2 r + µ 0 i d/2 ( ) µ0 i dr +2 2π(d r) a 2πr + µ 0 i dr 2π(d r) ( ( )) d a 1 2ln + µ ( ) 0i d a d π ln a where the first term is the flux within the wires and will be neglected (as the problem suggests). Thus, the flux is approximately Φ B µ 0 il/π ln((d a)/a). Now, we use Eq (with N =1)toobtain the inductance: L = Φ B = µ ( ) 0l d a i π ln. a

6 40. (a) Speaking anthropomorphically, the coil wants to fight the changes so if it wants to push current rightward (when the current is already going rightward) then i must be in the process of decreasing. (b) From Eq (in absolute value) we get L = E di/dt = 17 V 2.5kA/s = H.

7 54. (a) When switch S is just closed (case I), V 1 = E and i 1 = E/R 1 =10V/5.0Ω = 2.0A. After a long time (case II) we still have V 1 = E, soi 1 =2.0A. (b) Case I: since now E L = E, i 2 = 0; case II: since now E L =0,i 2 = E/R 2 =10V/10 Ω = 1.0A. (c) Case I: i = i 1 + i 2 =2.0A+0=2.0A; case II: i = i 1 + i 2 =2.0A+1.0A=3.0A. (d) Case I: since E L = E, V 2 = E E L = 0; case II: since E L =0,V 2 = E E L = E =10V. (e) Case I: E L = E =10V; caseii:e L =0. (f) Case I: di 2 /dt = E L /L = E/L =10V/5.0H=2.0A/s; case II: di 2 /dt = E L /L =0.

8 64. We use 1 ly = m, and use the symbol V for volume. U B = Vu B = VB2 = ( m) 3 ( T) 2 2µ 0 2(4π 10 7 = J. H/m)

9 72. The coil-solenoid mutual inductance is M = M cs = NΦ cs i s = N(µ 0i s nπr 2 ) i s = µ 0 πr 2 nn. As long as the magnetic field of the solenoid is entirely contained within the cross-section of the coil we have Φ sc = B s A s = B s πr 2, regardless of the shape, size, or possible lack of close-packing of the coil.

Chapter 30 Inductance

Chapter 30 Inductance In this chapter we investigate the properties of an inductor in a circuit. There are two kinds of inductance mutual inductance and self-inductance. An inductor is formed by taken