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Chapter 27: Problem 43

A generator consists of a rectangular coil \(75 \mathrm{cm}\) by \(1.3 \mathrm{m},\) spinning in a 0.14 -T magnetic field. If it's to produce a \(60-\mathrm{Hz}\) alternating emf with peak value \(6.7 \mathrm{kV},\) how many turns must it have?

Short Answer

Expert verified
The generator coil must have approximately 796 turns.

Step by step solution

01

Understand the Problem and Identify Known Quantities

The problem gives us the dimensions of the coil, the magnetic field strength (B = 0.14 T), the frequency of the alternating emf (f = 60 Hz) and the peak value of the emf (E_peak = 6.7 kV = 6700 V). We are tasked with finding the number of turns in the coil (N).
02

Identify the Relevant Formulas

To solve this problem, we will use Faraday's law of electromagnetic induction, which states that the magnitude of the induced emf in a coil is proportional to the rate of change of magnetic flux. The formula is \(E_peak = NABf\), where A is the area of the coil, B is the magnetic field strength, f is the frequency, and N is the number of turns.
03

Calculate the Area of the Coil

First, calculate the area of the rectangular coil using the formula \(A = width × height = 75 cm × 1.3 m = 0.975 m²\). Note that we converted cm to m to keep the units consistent.
04

Solve for the Number of Turns

Rearrange Faraday's law to solve for N: \(N = E_peak / ABf\). Substitute the known values and compute: \(N = 6700 V / (0.14 T × 0.975 m² × 60 s⁻¹) = approximately 796 turns\).
05

Round to the Nearest Whole Number

Since the number of turns in a coil cannot be fractional, we round 796 to the nearest whole number. Thus, the coil must have approximately 796 turns.

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Most popular questions from this chapter

The current in a series \(R L\) circuit increases to \(20 \%\) of its final value in \(3.1 \mu \mathrm{s}\). If \(L=1.8 \mathrm{mH},\) what's the resistance?

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