Homework 10 Solutions

Thus t equivalent to 2ft if is understood to be in

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Unformatted text preview: of cos(2πft) or even cos(2πft + 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 affect whether 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 = 2πft) 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 ) = − NBA = N Bab 2π f sin ( 2π ft ) ε = −N 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 = 2πf NabB. (b) We solve ε0 = 150 V = 2π f NabB 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 m2. 12. To have an induced emf, the magnetic field must be perpendicular (or have a nonzero component perpendicular) to the coil, and must be changing with time. ˆ (a) For B = (4.00 ×10−2 T/m) yk , dB / dt = 0 and hence ε = 0. (b) None. ˆ (c) For B = (6.00 ×10−2 T/s)tk , dΦB dB ε = − dt = −A dt = −(0.400 m × 0.250 m)(0.0600 T/s) = −6.00 mV, or |ε| = 6.00 mV. 1168 CHAPTER 30 (d) Clockwise. ˆ (e) For B = (8.00 ×10−2 T/m ⋅ s) yt k , ΦB = (0.400)(0.0800t) ∫ ydy = 1.00 ×10−3 t , in SI units. The induced emf is ε = − d ΦB / dt = −1.00 mV, or |ε| = 1.00 mV. (f) Clockwise. (g) Φ B = 0 ⇒ ε = 0 . (h) None. (i) Φ B = 0 ⇒ ε = 0 . (j) None. 13. The amount of charge is q (t ) = 1 A 1.20 × 10−3 m 2 [Φ B (0) − Φ B (t )] = [ B(0) − B(t )] = [1.60 T − ( − 1.60 T)] 13.0 Ω R R = 2.95 × 10−2 C . 14. Figure 30-40(b) demonstrates that dB / dt (the slope of that line) is 0.003 T/s. Thus, in absolute value, Faraday’s law becomes ε =− dΦB d ( BA) dB =− = −A dt dt dt where A = 8 ×10−4 m2. We related the induced emf to resistance and current using Ohm’s law. The current is estimated from Fig. 30-40(c) to be i = dq / dt = 0.002 A (the slope of that line). Therefore, the resistance of the loop is | ε | A | dB / dt | (8.0 × 10−4 m 2 )(0.0030 T/s) R= = = = 0.0012 Ω . i i 0.0020 A 15. (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 1169 εi = − dΦB L2 dB . =− 2 dt dt Now B = 0.042 – 0.870t and dB/dt = –0.870 T/s. Thus, εi = (2.00 m) 2 (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...
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This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

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