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P2213_HW13_solns_Fall2008

# P2213_HW13_solns_Fall2008 - Physics 2213 HW#13 Solutions...

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- 1 - Physics 2213 HW #13 — Solutions Fall 2008 29.8 [Loop & Changing B] Apply Faraday’s law. Let A be upward in Figure 29.28 in the textbook. (a) ind ( ) B d d B A dt dt Φ = = E = ( ) 1 1 (0.057s ) 2 1 (0.057s ) ind sin60 sin60 (1.4 T) ( )(sin60 )(1.4 T)(0.057 s ) t t dB d A A e r e dt dt π = ° = ° = ° E 1 1 2 1 (0.057s ) (0.057 s ) ind (0.75 m) (sin60 )(1.4 T)(0.057 s ) (0.12 V) . t t e e π = ° = E (b) 1 1 0 10 10 (0.12 V). = = E E 1 (0.057 s ) 1 10 (0.12 V) (0.12 V) . t e = 1 ln(1/10) (0.057 s ) t = and 40.4 s. t = (c) B is in the direction of A so B Φ is positive. B is getting weaker, so the magnitude of the flux is decreasing and / 0. B d dt Φ < Faraday’s law therefore says 0. > E Since 0, > E the induced current must flow counterclockwise as viewed from above. Recap: The flux changes because the magnitude of the magnetic field is changing. 29.20 [Slide-Rod Circuit] Use the results of Example 29.6. Use the three approaches specified in the problem for determining the direction of the induced current. / I R = E . Let A be directed into the figure, so a clockwise emf is positive. (a) (5.0 m/s)(0.750 T)(1.50 m) 5.6 V vBl = = = E (b) (i) Let q be a positive charge in the moving bar, as shown in Figure (a) at right. The magnetic force on this charge is q × F = v B , which points upward . This force pushes the current in a counterclockwise direction through the circuit. (ii) B Φ is positive and is increasing in magnitude, so / 0 B d dt Φ > . Then by Faraday’s law 0 < E and the emf and induced current are counterclockwise. (iii) The flux through the circuit is increasing, so the induced current must cause a magnetic field out of the paper to oppose this increase. Hence this current must flow in a counterclockwise sense, as shown in Figure (b) at right. (c) . RI = E 5.6 V 0.22 A. 25 I R = = = Ω E All three methods agree on the direction of the induced current. 29.26 [Moving Square Loop] Use Faraday’s law to calculate the induced emf. Ohm’s law applied to the loop gives I . Use Eq.(27.19) to calculate the force exerted on each side of the loop. The loop before it starts to enter the magnetic field region is sketched above right. For 3 /2 or 3 /2 x L x L < > the loop is completely outside the field region.

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P2213_HW13_solns_Fall2008 - Physics 2213 HW#13 Solutions...

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