27_InstSolManual_PC

# 27_InstSolManual_PC - ELECTROMAGNETIC INDUCTION E XERCISES...

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Unformatted text preview: ELECTROMAGNETIC INDUCTION E XERCISES Sections 27.2 Faraday’s Law and 27.3 Induction and Energy 15. I NTERPRET In this problem we are asked to verify that the SI unit of the rate of change of magnetic flux is volt. D EVELOP We first note that the left-hand-side of Equation 27.2, / , B d dt ε = - Φ represents the induced emf which has units of volt. From the definition of magnetic flux given in Equation 27.1a, we see that it has SI units of 2 T m . ⋅ E VALUATE The reasoning above shows that the units of / B d dt Φ are 2 2 T m /s (N/A m)(m /s) (N m/A s) J/C V ⋅ = ⋅ = ⋅ ⋅ = = A SSESS Faraday’s law relates the induced emf to the change in flux. It is the rate of change of flux, and not the flux or the magnetic field that gives rise to an induced emf. 16. For a stationary plane loop in a uniform magnetic field, the integral for the flux in Equation 27.1a is just B B A φ = ⋅ = r r 2 4 cos (80 mT) (2.5 cm) cos30 1.36 10 Wb. BA θ π- = ° = × (The SI unit of flux, 2 T m , ⋅ is also called a weber, Wb.) 17. I NTERPRET This problem is about the rate of change of magnetic flux through a loop due to a changing magnetic field. D EVELOP For a stationary plane loop in a uniform magnetic field, the magnetic flux is given by Equation 27.1b, cos . B B A BA θ Φ = ⋅ = r r Note that the SI unit of flux, 2 T m , ⋅ is also called a weber, Wb. The rate of change of magnetic flux is / / . B B d dt t Φ = ∆Φ ∆ E VALUATE (a) The magnetic field at the beginning 1 ( 0) t = is 2 2 4 1 1 1 1 1 (40 cm) (5 mT) 6.28 10 Wb 4 4 B A d B π π- Φ = = = = × (b) The magnetic field at 2 25 ms t = is 2 2 3 2 2 2 1 1 (40 cm) (55 mT) 6.91 10 Wb 4 4 B A d B π π- Φ = = = = × (c) Since the field increases linearly, the rate of change of magnetic flux is 3 3 2 1 2 1 6.91 10 Wb 0.628 10 Wb 0.251 V 25 ms B B d dt t t t-- Φ ∆Φ Φ - Φ ×- × = = = = ∆- From Faraday’s law, this is equal to the magnitude of the induced emf, which causes a current | | 0.251 V = 2.51 mA 100 I R ε = = Ω in the loop. sol coil sol sol coil sol | | ( ) B dB d d N B A N A dt dt dt ε Φ = = = (d) The direction must oppose the increase of the external field downward, hence the induced field is upward and I is CCW when viewed from above the loop. A SSESS Since / ( / ) B t B t A ∆Φ ∆ = ∆ ∆ with the area of the loop kept fixed, the induced emf and hence the current scale linearly with the value / . B t ∆ ∆ 27.1 27 27.2 Chapter 27 18. The flux through a stationary loop perpendicular to a magnetic field is B BA φ = (see Exercise 16), so Faraday’s law (Equation 27.2) and Ohm’s law (Equation 24.5) relate this to the magnitude of the induced current: | / | I R ε = = 2 | / |/ | / |/ . Therefore | / | / (320 mA)(12 )/(240 cm ) 160 T/s....
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27_InstSolManual_PC - ELECTROMAGNETIC INDUCTION E XERCISES...

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