30_InstructorSolutions

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

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30-1 I NDUCTANCE 30.1. IDENTIFY and SET UP: Apply Eq.(30.4). EXECUTE: (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 EVALUATE: The induced emf is the same in either case. A constant / di dt produces a constant emf. 30.2. IDENTIFY: 2 1 i M t Δ = Δ E and 1 2 . i M t Δ = Δ E 22 1 B N M i Φ = , where 2 B Φ is the flux through one turn of the second coil. SET UP: M is the same whether we consider an emf induced in coil 1 or in coil 2. EXECUTE: (a) 3 3 2 1 1.65 10 V 6.82 10 H 6.82 mH / 0.242 A/s M it × = × = ΔΔ E (b) 3 4 1 2 2 (6.82 10 H)(1.20 A) 3.27 10 Wb 25 B Mi N × Φ= = = × (c) 33 2 1 (6.82 10 H)(0.360 A/s) 2.46 10 V 2.46 mV i M t −− Δ × = × = Δ E EVALUATE: 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. IDENTIFY: Replace units of Wb, A and Ω by their equivalents. SET UP: 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 . EXECUTE: 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. = = ⋅= = = = EVALUATE: We may use whichever equivalent unit is the most convenient in a particular problem. 30.4. IDENTIFY: Changing flux from one object induces an emf in another object. (a) SET UP: The magnetic field due to a solenoid is 0 . B nI μ = EXECUTE: 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 ( ) 42 8 1 1.81 10 T (0.0100 m) = 5.68 10 Wb B BA Φ= = × × (b) SET UP: The flux is the same through each turn of both solenoids due to the geometry, so 2, 2 1 11 BB NN M ii ΦΦ EXECUTE: ( ) 8 5 (25) 5.68 10 Wb 1.18 10 H 0.120 A M × × (c) SET UP: The induced emf is 1 2 . di M dt =− E EXECUTE: ( ) 5 2 1.18 10 H (1750 A/s) 0.0207 V × E EVALUATE: A mutual inductance around 5 10 H is not unreasonable. 30

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30-2 Chapter 30 30.5. IDENTIFY and SET UP: Apply Eq.(30.5). EXECUTE: (a) () 22 1 400 0.0320 Wb 1.96 H 6.52 A B N M i Φ == = (b) 3 11 2 1 21 (1.96 H)(2.54 A) so 7.11 10 Wb 700 B B NM i M iN Φ = = = × EVALUATE: 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. IDENTIFY: A changing current in an inductor induces an emf in it. (a) SET UP: The self-inductance of a toroidal solenoid is 2 0 . 2 N A L r μ π = EXECUTE: 72 4 2 4 (4 10 T m/A)(500) (6.25 10 m ) 7.81 10 H 2 (0.0400 m) L −− ×⋅ × × (b) SET UP: The magnitude of the induced emf is . di L dt = E EXECUTE: 4 3 5.00 A 2.00 A 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. EVALUATE: This is a reasonable value for self-inductance, in the range of a mH.
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30_InstructorSolutions - INDUCTANCE 30 Apply Eq.(30.4). di...

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