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P2213_HW14_solns_Fall2008

# P2213_HW14_solns_Fall2008 - Physics 2213 HW#14 Solutions...

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- 1 - Physics 2213 HW #14 — Solutions Fall 2008 30.6 [Inductor] A changing current in an inductor induces an emf in it. (a) The self-inductance of a toroidal solenoid is 2 0 . 2 N A L r μ π = 7 2 4 2 4 (4 10 T m/A)(500) (6.25 10 m ) 7.81 10 H 2 (0.0400 m) L π π - - - × × = = × This is a reasonable value for self-inductance, in the range of a mH. (b) The magnitude of the induced emf is . di L dt = E ( 29 4 3 5.00 A 2.00 A 7.81 10 H 0.781 V 3.00 10 s - - - = × = × E (c) The current is decreasing, so the induced emf will be in the same direction as the current, which is from a to b, making b at a higher potential than a. 30.19 [LR Circuit] Apply Kirchhoff’s loop rule to the circuit. i ( t ) is given by Eq.(30.14). di dt is positive as the current increases from its initial value of zero. 0 R L v v - - = E ( 29 ( / ) 0 so 1 R L t di iR L i e dt R - - - = = - E E (a) Initially ( t = 0), i = 0 so 0 di L dt - = E 6.00 V 2.40 A/s 2.50 H di dt L = = = E (b) 0 di iR L dt - - = E (Use this equation rather than Eq.(30.15) since i rather than t is given.) Thus 6.00 V (0.500 A)(8.00 ) 0.800 A/s 2.50 H di iR dt L - - = = = E (c) ( 29 ( 29 ( / ) (8.00 /2.50 H)(0.250 s) 0.800 6.00 V 1 1 0.750 A(1 ) 0.413 A 8.00 R L t i e e e R - - - = - = - = - = E (d) Final steady state means and 0, so 0. di t iR dt → ∞ - = E 6.00 V 0.750 A 8.00 i R = = = E Our results agree with Fig.30.12 in the textbook. The current is initially zero and increases to its final value of / . R E The slope of the current in the figure, which is di/dt , decreases with t .

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- 2 - 30.43 [Mutual Inductance] The emf 2 E in solenoid 2 produced by changing current 1 i in solenoid 1 is given by 1 2 . i M t = E The mutual inductance of two solenoids is derived in Example 30.1. For the two solenoids in this problem 0 1 2 , AN N M l μ = where A is the cross-sectional area of the inner solenoid and l is the length of the outer solenoid.
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