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Unformatted text preview: SOLUTIONS, ASSIGNMENT 26 29.28. I DENTIFY : Use Eq.(29.10) to calculate the induced electric field E at a distance r from the center of the solenoid. Away from the ends of the solenoid, B nI μ = inside and B = 0 outside. (a) S ET U P : The end view of the solenoid is sketched in Figure 29.28. Let R be the radius of the solenoid. Figure 29.28 Apply B d d dt Φ ⋅ E l = r r ú to an integration path that is a circle of radius r , where r < R . We need to calculate just the magnitude of E so we can take absolute values. E XECUTE : (2 ) d E r π ⋅ = E l r r ú 2 2 , B B d dB B r r dt dt π π Φ Φ = = 2 implies (2 ) B d dB d E r r dt dt π π Φ ⋅ =  = E l r r ú 1 2 dB E r dt = , so dB dI B nI n dt dt μ μ = = Thus 7 1 4 1 1 2 2 (0.00500 m)(4 10 T m/A)(900 m )(60.0 A/s) 1.70 10 V/m dI E r n dt μ π = = × ⋅ = × (b) r = 0.0100 cm is still inside the solenoid so the expression in part (a) applies. 7 1 4 1 1 2 2 (0.0100 m)(4 10 T m/A)(900 m )(60.0 A/s) 3.39 10 V/m dI E r n dt μ π = = × ⋅ = × E VALUATE : Inside the solenoid E is proportional to r , so E doubles when r doubles. 29.30. I DENTIFY : Use Eq.(29.10) to calculate the induced electric field E and use this E in Eq.(29.9) to calculate E between two points. (a) S ET U P : Because of the axial symmetry and the absence of any electric charge, the field lines are concentric circles. (b) See Figure 29.30. E r is tangent to the ring. The direction of E r (clockwise or counterclockwise) is the direction in which current will be induced in the ring. Figure 29.30 E XECUTE : Use the sign convention for Faraday’s law to deduce this direction. Let A r be into the paper. Then B Φ is positive. B decreasing then means B d dt Φ is negative, so by , B d dt Φ =  E E is positive and therefore clockwise. Thus E r is clockwise around the ring. To calculate E apply B d d dt Φ ⋅ E l = r r...
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This note was uploaded on 03/11/2010 for the course PHYSICS 1523 taught by Professor Dfs during the Fall '10 term at Dallas.
 Fall '10
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