Ch30Word - Chapter 30, p. 1 CHAPTER 30 Inductance; and...

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Chapter 30, p. 1 CHAPTER 30 – Inductance; and Electromagnetic Oscillations 1. The magnetic field of the long solenoid is essentially zero outside the solenoid. Thus there will be the same linkage of flux with the second coil and the mutual inductance will be the same: M = µ 0 N 1 N 2 A / ¬ . 2. ( a ) We find the mutual inductance from M = N 1 N 2 A / ¬ = (2000)(4p × 10 –7 T · m/A)(300 turns)(100 turns)p(0.0200 m) 2 /(2.44 m) = 3.88 × 10 –2 H = 38.8 mH . ( b ) The induced emf in the second coil is å = – M ? I 1 /? t = ( 3 . 8 8 × 10 –2 H)(0 – 12.0 A)/(0.0980 s) = 4.75 V . 3. The magnetic field inside the outer solenoid is B = 0 n 1 I 1 . The magnetic flux linked with the inner solenoid is Φ 21 = 0 n 1 I 1 A 2 . Thus the mutual inductance with the N 2 turns of the inner solenoid is M 21 = N 2 Φ 21 / I 1 = n 2 ¬ 0 n 1 A 2 , so the mutual inductance per unit length is M / ¬ = 0 n 1 n 2 p r 2 2 . 4. We find the mutual inductance of the system by finding the mutual inductance of the small coil. The magnetic field inside the long solenoid is along the axis with a magnitude B 1 = 0 n 1 I 1 = 0 N 1 I 1 / ¬ . We find the magnetic flux through the coil by using the area perpendicular to the field: Φ 21 = ( 0 N 1 I 1 / ¬ ) A 2 sin θ . Thus the mutual inductance with the N 2 loops of the coil is M 21 = N 2 Φ 21 / I 1 = ( 0 N 1 N 2 A 2 / ¬ ) sin . 5. We find the mutual inductance of the system by finding the mutual inductance of the loop. The magnetic field of the long wire depends only on the distance from the wire. To find the magnetic flux through the loop, we choose a strip a distance x from the wire with width d x : Φ Β = d A = 0 I 2 π x w d x 1 2 = 0 Iw 2 π ln 2 1 . The mutual inductance is M = Φ B / I = ( 0 w /2p) ln( ¬ 2 / ¬ 1 ). 6. We find the induced emf from å = – L ? I /? t = – (180 mH)(38.0 mA – 20.0 mA)/(340 ms) = – 9.53 mV . The emf is opposite to the direction of the current, to oppose the increase in the current. 7. We estimate the inductance by using the inductance of a solenoid: L = 0 N 2 A / ¬ = (4p × 10 –7 T · m/A)(20,000 turns) 2 p(1.85 × 10 –2 m) 2 /(0.45 m) = 1.2 H . 8. Because the current in increasing, the emf is negative. We find the self-inductance from å = – L ? I /? t ; – 8.50 V = – L [23.0 mA – (– 22.0 mA)]/(21.0 ms), which gives L = 3.97 H . w x d x I ¬ 1 B ¬ 2
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Chapter 30, p. 2 9. We use the result from Ex. 30–4: L = ( µ 0 ¬ /2p) ln( r 2 / r 1 ) = ( 0 ¬ /2p) ln( D 2 / D 1 ) = ( 2 × 10 –7 T · m/A)(22.0 m) ln(3.5 mm/2.0 mm) = 2.5 × 10 –6 H . 10. We find the current from å = – L ? I /? t ; – 35 V = – (150 mH)( I 0 – 0)/(3.0 ms), which gives I 0 = 0.70 A . 11. We use the result for the inductance from Ex. 30–4: L / ¬ = ( 0 /2p) ln( r 2 / r 1 ); ( 2 × 10 –7 T · m/A) ln(3.0 mm/ r 1 ) = 40 × 10 –9 H/m, which gives r 1 = 2.5 mm . 12. ( a ) The magnetic flux through coil 1 is due to its own current and the current in the other coil. From the definition of self-inductance, the flux from its own current is Φ 11 = L 1 I 1 . From the definition of mutual inductance, the flux from the current in the other coil is Φ 12 = M 12 I 2 .
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch30Word - Chapter 30, p. 1 CHAPTER 30 Inductance; and...

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