A car alternator produces peak output Vo to charge the car's 12 V battery. Suppose the alternator coil has diameter d and is spinning at 1200 revolutions per minute in a magnetic field B. How many turns of wire are required to produce peak output Vo? Vmax = 14.3 V; diameter = 15 cm; revolution frequency = 1270 revolutions per minute; B = 0.16 T

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A car alternator produces peak output Vo to charge the
car's 12 V battery. Suppose the alternator coil has
diameter d and is spinning at 1200 revolutions per
minute in a magnetic field B. How many turns of wire
are required to produce peak output Vo?
Vmax = 14.3 V; diameter = 15 cm; revolution frequency = 1270 revolutions per minute; B = 0.16 T
Transcribed Image Text:A car alternator produces peak output Vo to charge the car's 12 V battery. Suppose the alternator coil has diameter d and is spinning at 1200 revolutions per minute in a magnetic field B. How many turns of wire are required to produce peak output Vo? Vmax = 14.3 V; diameter = 15 cm; revolution frequency = 1270 revolutions per minute; B = 0.16 T
Expert Solution
Step 1: INTRODUCTION

Given Data :

The maximum voltage output is V subscript m a x end subscript equals 14.3 space V

The diameter of coil is d equals 15 space c m space equals 0.15 space m

Revolution frequency is omega equals 1270 space fraction numerator r e v over denominator m i n end fraction equals fraction numerator 1270 cross times 2 pi over denominator 60 end fraction fraction numerator r a d over denominator s end fraction equals 133 space fraction numerator r a d over denominator s end fraction

The magnitude of the magnetic field is B equals 0.16 space T


Now, Area of the coil is 

A equals pi d squared over 4
space space space equals pi left parenthesis 0.15 right parenthesis squared over 4
space space space equals space 0.01767 space m squared


Voltage is given by

V subscript m a x end subscript equals N B A omega..... left parenthesis 1 right parenthesis

Where,

N is the number of turns .


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