Lesson_4.4_Homework_and_Solutions

# Lesson_4.4_Homework_and_Solutions - Lesson 4.4 Homework...

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Unformatted text preview: Lesson 4.4 Homework Solutions 1. A 500-turn coil has an area of 5.5 cm 2. If it is rotated in a uniform magnetic field of 0.8 T, (i) what frequency of rotation will generate a maximum emf of 10 V in the coil? (ii) If the coil rotates at 60 Hz, what is the maximum induced emf? Solution N = 500 turns, A = 5.5 × 10-4 m 2 , E max = 10 V, B = 0.8 T E max = NAB ϖ max 4 2 1 10 ( ) 500 5.5 45.5 10 0.8 E V i NAB m s T ϖ-- = = = × × × (ii) ϖ = 2 π f 1 45.5 2 2 7.2 s H f z ϖ π π- = = = 2. A 700-turns rectangular coil of area 3 cm 2 rotates in a magnetic field of 9000 G. (i) What is the maximum emf generated when the coil rotates 60 times a second? (ii) What must be the frequency of rotation to generate a maximum emf of 120 V? Solution N = 700 turns, A = 3.0 × 10-4 m 2 , B = 9000 G = 0.9 T (i) f = 60 Hz, ϖ = 2 π f = 2 π × 60 Hz = 377 Hz E max = NAB ϖ = 700 × 3.0 × 10-4 m 2 × 0.9 T × 377 Hz = 71.3 V (ii) ϖ = 2 π f , & max = 120 V E max = NAB ϖ max 4 2 120 700 3.0 10 3 . 6 5 9 E V NAB m Hz T ϖ- = = = × × × 1 635 2 1 2 01 s f Hz ϖ π π- = = = 3. A 60-W light bulb is plugged into 110-V supply. Find (i) I rms , (ii) I max and (iii) the maximum power dissipated in the bulb. Solution P ave = 60 W, Vrms = 110 V (i) P ave = I rms × V rms 0. 6 1 55 1 0 ave rms rms W I V A P V = = = (ii) I max = √ 2 I rms = √ 2 × 0.55 A = 1.05 A (ii) V max = √ 2 × V rms = √ 2 × 110 V = 155.6 V P max = I max V max = 1.05 A × 155.6 V = 163.6 W 4. A 60-Hz generator producing a maximum emf of 12.0 V is connected across a 3.0- Ω resistor. (i) What is the angular frequency resistor....
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Lesson_4.4_Homework_and_Solutions - Lesson 4.4 Homework...

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