30_InstructorSolutions - INDUCTANCE 30 Apply Eq(30.4 di(a...

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30-1 I NDUCTANCE 30.1. I DENTIFY and S ET U P : Apply Eq.(30.4). E XECUTE : (a) 4 1 2 (3.25 10 H)(830 A/s) 0.270 V; di M dt = = × = E yes, it is constant. (b) 2 1 ; di M dt = E M is a property of the pair of coils so is the same as in part (a). Thus 1 0.270 V. = E E VALUATE : The induced emf is the same in either case. A constant / di dt produces a constant emf. 30.2. I DENTIFY : 2 1 i M t Δ = Δ E and 1 2 . i M t Δ = Δ E 2 2 1 B N M i Φ = , where 2 B Φ is the flux through one turn of the second coil. S ET U P : M is the same whether we consider an emf induced in coil 1 or in coil 2. E XECUTE : (a) 3 3 2 1 1.65 10 V 6.82 10 H 6.82 mH / 0.242 A/s M i t × = = = × = Δ Δ E (b) 3 4 1 2 2 (6.82 10 H)(1.20 A) 3.27 10 Wb 25 B Mi N × Φ = = = × (c) 3 3 2 1 (6.82 10 H)(0.360 A/s) 2.46 10 V 2.46 mV i M t Δ = = × = × = Δ E E VALUATE : We can express M either in terms of the total flux through one coil produced by a current in the other coil, or in terms of the emf induced in one coil by a changing current in the other coil. 30.3. I DENTIFY : Replace units of Wb, A and Ω by their equivalents. S ET U P : 2 1 Wb 1 T m . = 1 T 1 N/(A m). = 1 N m 1 J. = 1 A 1 C/s. = 1 V 1 J/C. = 1 V/A 1 . = Ω E XECUTE : 2 2 2 1H 1 Wb/A 1T m /A 1 N m/A 1J/A 1(J/[A C])s 1(V/A)s 1 s. = = = = = = = E VALUATE : We may use whichever equivalent unit is the most convenient in a particular problem. 30.4. I DENTIFY : Changing flux from one object induces an emf in another object. (a) S ET U P : The magnetic field due to a solenoid is 0 . B nI μ = E XECUTE : The above formula gives ( ) 7 4 1 4 10 T m/A (300)(0.120 A) =1.81 10 T 0.250 m B π × = × The average flux through each turn of the inner solenoid is therefore ( ) 4 2 8 1 1.81 10 T (0.0100 m) = 5.68 10 Wb B B A π Φ = = × × (b) S ET U P : The flux is the same through each turn of both solenoids due to the geometry, so 2 ,2 2 ,1 1 1 B B N N M i i Φ Φ = = E XECUTE : ( ) 8 5 (25) 5.68 10 Wb 1.18 10 H 0.120 A M × = = × (c) S ET U P : The induced emf is 1 2 . di M dt = − E E XECUTE : ( ) 5 2 1.18 10 H (1750 A/s) 0.0207 V = − × = − E E VALUATE : A mutual inductance around 5 10 H is not unreasonable. 30
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30-2 Chapter 30 30.5. I DENTIFY and S ET U P : Apply Eq.(30.5). E XECUTE : (a) ( ) 2 2 1 400 0.0320 Wb 1.96 H 6.52 A B N M i Φ = = = (b) 3 1 1 2 1 2 1 (1.96 H)(2.54 A) so 7.11 10 Wb 700 B B N Mi M i N Φ = Φ = = = × E VALUATE : M relates the current in one coil to the flux through the other coil. Eq.(30.5) shows that M is the same for a pair of coils, no matter which one has the current and which one has the flux. 30.6. I DENTIFY : A changing current in an inductor induces an emf in it. (a) S ET U P : The self-inductance of a toroidal solenoid is 2 0 . 2 N A L r μ π = E XECUTE : 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 × × = = × (b) S ET U P : The magnitude of the induced emf is . di L dt = E E XECUTE : ( ) 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.
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