ε m = Aµ 0 ni0ω = (6.8 ×10−6 m 2 )(4π × 10 −7 T ⋅ m A)(85400 / m)(1.28 A)(212 rad/s)
= 1.98 ×10−4 V.
2. (a) ε =
dΦ B d = 6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV. dt dt
(b) Appealing to Lenz’s law (especially Fig. 30-5(a)) we see that the current flow in the loop is clockwise. Thus, the current is to leftthrough R.
3. (a) We use ε = –dΦB/dt = –πr2dB/dt. For 0 < t < 2.0 s:
ε = −πr 2
dB 2 0.5T = −π ( 0.12m ) = −1.1×10−2 V. dt 2.0s
(b) 2.0 s < t < 4.0 s: ε ∝ dB/dt = 0. (c) 4.0 s < t < 6.0 s:
ε = − πr 2
dB = − π 012 m . dt
. g FGH 6.0−s0−.54T.0sIJK = 11 × 10
4. The resistance of the loop is
π ( 0.10 m ) L = (1.69 × 10−8 Ω ⋅ m ) = 1.1× 10−3 Ω. −3 2 Aπ ( 2.5 × 10 ) / 4
We use i = |ε|/R = |dΦB/dt|/R = (πr2/R)|dB/dt|. Thus
−3 dB iR (10 A ) (1.1×10 Ω ) = = = 1.4 T s. 2 dt π r 2 π ( 0.05 m )
5. The total induced emf is given by
ε = −N
dΦB dB d di di = − NA = − NA ( µ 0 ni ) = − N µ 0 nA = − N µ 0 n(π r 2 ) dt dt dt dt dt
= −(120)(4π ×10 −7 T ⋅ m A)(22000/m) π ( 0.016m ) = 0.16V.
1.5 A 0.025 s
Ohm’s law then yields i =| ε |/ R = 0.016 V / 5.3Ω = 0.030 A .
6. Using Faraday’s law, the induced emf is d ( πr 2 ) d ( BA ) dΦB dA dr =− = −B = −B = −2πrB ε =− dt dt dt dt dt = −2π ( 0.12m )( 0.800T )( −0.750m/s ) = 0.452V.
7. The flux Φ B = BA cosθ does not change as the loop is rotated. Faraday’s law only leads to a nonzero induced emf when the flux is changing, so the result in this instance is 0.
8. Thefield (due to the current in the straight wire) is out-of-the-page in the upper half of the circle and is into the page in the lower half of the circle, producing zero net flux, at any time. There is no induced current in the circle.
9. (a) Let L be the length of a side of the square circuit. Then the magnetic flux through the circuit is Φ B = L2 B / 2 , and the induced emf is
εi = −
dΦB L2dB =− . dt 2 dt
Now B = 0.042 – 0.870t and dB/dt = –0.870 T/s. Thus, (2.00 m) 2 εi = (0.870 T / s) = 1.74 V. 2 The magnetic field is out of the page and decreasing so the induced emf is counterclockwise around the circuit, in the same direction as the emf of the battery. The total emf is
ε + εi = 20.0 V + 1.74 V = 21.7 V.
(b) The current is in the sense of the total emf (counterclockwise).dB 10. Fig. 30-41(b) demonstrates that dt (the slope of that line) is 0.003 T/s. Thus, in absolute value, Faraday’s law becomes
ε = − dt = − A dt
where A = 8 ×10−4 m2. We related the induced emf to resistance and current using Ohm’s dq law. The current is estimated from Fig. 30-41(c) to be i = dt = 0.002 A (the slope of that line). Therefore, the resistance of the loop is R= |ε | / i =
(8 x 10-4 )(0.003) = 0.0012 Ω . 0.002
11. (a) It should be emphasized that the result, given in terms of sin(2π ft), could as easily be given in terms of cos(2π ft) or even cos(2π ft + φ) where φ is a phase constant as discussed in Chapter 15. The angular position θ of the rotating coil is measured from some reference line (or plane), and which line one chooses will affectwhether the magnetic flux should be written as BA cosθ, BA sinθ or BA cos(θ + φ). Here our choice is such that Φ B = BA cosθ . Since the coil is rotating steadily, θ increases linearly with time. Thus, θ = ωt (equivalent to θ = 2π ft) if θ is understood to be in radians (and ω would be the angular velocity ). Since the area of the rectangular coil is A=ab , Faraday’s law leads to d ( BA cos θ ) d cos (2π ft ) ε = −N = − NBA = N Bab 2π f sin ( 2π ft ) dt dt which is the desired result, shown in the problem statement. The second way this is written (ε0 sin(2π ft)) is meant to emphasize that the voltage output is sinusoidal (in its time dependence) and has an amplitude of ε0 = 2π f N abB. (b) We solve ε0 = 150 V = 2π f N abB when f = 60.0 rev/s and B = 0.500 T. The three unknowns are N, a, and...