Calculating Magnetic Field from Generator Graph

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SUMMARY

The discussion focuses on calculating the magnetic field from the electromotive force (emf) of a generator using the formula B = emf / (-NAωsin(ωt)). The user provided specific values: a cross-sectional area of 0.03 m², 160 turns, a frequency of 2.4 Hz, and an angular speed of 14.96 rad/s. The calculated magnetic field value is approximately 0.38993 T, derived from substituting the emf value of 28 V into the rearranged formula. The approach and calculations presented are accurate and confirm the relationship between emf and magnetic flux change.

PREREQUISITES
  • Understanding of electromagnetic induction principles
  • Familiarity with the formula for emf in generators
  • Knowledge of trigonometric functions, particularly sine
  • Basic proficiency in calculus for differentiation
NEXT STEPS
  • Explore the derivation of Faraday's Law of Electromagnetic Induction
  • Learn about the relationship between frequency and angular speed in generators
  • Investigate the effects of varying magnetic fields on emf generation
  • Study practical applications of magnetic field calculations in electrical engineering
USEFUL FOR

This discussion is beneficial for physics students, electrical engineering students, and professionals involved in generator design and analysis, particularly those focusing on electromagnetic principles and calculations.

candyq27
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Homework Statement


I have a graph of an emf of a generator vs time, and i know the cross-sectional area per turn is 0.03m^2 and there are 160 turns. From the wave graph I figured out the frequency of the generator is 2.4Hz and the angular speed is 14.96rad/s. I need to figure out the magnetic field.


Homework Equations


e=-NAcos0B/t


The Attempt at a Solution



I found the frequency and angular speed from the graph but I have no idea how to find the magnetic field from a graph. Please help. Thank you.
 
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It's important to realize that the emf is directly proportional to the change of magnetic flux:

\Phi = BS\cos{\omega}t

emf = -N \frac{d\Phi}{dt},

that is:

emf = -NBS \frac{d}{dt}\cos{\omega}t = -NBS\omega\sin{\omega}t

Then rearrange it to get the formula for B.
 
Last edited:
Ok so I rearranged the formula to get B=emf/-NAwsinwt
= 28/(-160)(.03)(14.96)(sin(14.96*.315))
=28/71.808
=0.38993
Is that correct? Thanks!
 

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