Calculating Magnetism on Child Homework

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Homework Help Overview

The discussion revolves around calculating the maximum possible magnetic flux density near a child sleeping close to household wiring, given a specific power transmission and voltage. The problem involves concepts from electromagnetism and alternating current (AC) theory.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between root mean square (r.m.s) voltage and actual voltage, questioning how to convert between them. There are attempts to apply formulas related to magnetic flux density and current, with some confusion about the correct values to use.

Discussion Status

The discussion is active, with participants providing insights into the nature of AC voltage and current. Some have offered guidance on the formulas and concepts involved, while others express confusion about the definitions and calculations, particularly regarding the current values used in the problem.

Contextual Notes

Participants note discrepancies in the expected current values and question the assumptions made in the problem statement, particularly regarding the use of power in the context of AC circuits.

NewAgeBerean
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Homework Statement



A child sleeps at an average distance of 30cm from household wiring. The mains supply is 240V r.m.s. Calculate the max possible magnetic flux density in the region of child when the wire is transmitting 3.6 kW of power.

Thats the first part. I don't really understand what r.m.s is and how to convert it to normal V..


Homework Equations



VI=P
F=BIL

The Attempt at a Solution



B= (P U0) / (V 2pi r)
= (3.5 * 10^3)(4pi * 10^ -7) / (240 2pi (0.3))

Getting 9.7 * 10 - 2 - not correct.
 
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rms is just a way of stating the 'average' voltage

Shouldn't flux have units of field/area?
 
So how can u convert average voltage to actual voltage? on the working out they have a current of 21.2 A. How does that happen?

are my formulas wrong?
 
It's AC so the 'actual voltage' changes continually - it has an average (in simple terms) voltage of 240V so an average current of 3600W/240V = 15A

Now you need the equation for a field a certain distance from a current carrying wire.
 
Ok i use

B = (U0)(I)/ 2pi (0.3)
= (4.7*10^-7)(14.58) / 2pi (0.3)
= 3.64*10-6 T

They have 21.2 A. and have B= 1.5*10-5 T
 
Welcome to PF!

Hi NewAgeBerean! Welcome to PF! :wink:

For AC voltage …

Vrms = Vpeak/√2 = Vaverageπ/2√2 :smile:

(same with AC current … Irms = Ipeak/√2)
 
so in this case which V do i use - average or peak? Are my previous working out correct? and is the mark scheme answer to the question incorrect? (they used 21.2 A)
 
The question asks for the maximum possible, not the average. The the peak value is about 1.414 (square root of 2) times the rms (root mean square) value.

In real life the voltage in a circuit varies by plus or minus 10% from the nominal as well. Assuming the load's power remains constant, that could mean extra current if the voltage were low.
 
Looking at the question again, I'm confused by the use of "power" …
NewAgeBerean said:
Calculate the max possible magnetic flux density in the region of child when the wire is transmitting 3.6 kW of power.

… power varies with voltage in an AC current …

P = V2/R, and so Ppeak = Vpeak2/R, Paverage = Vrms2/R.

What exactly does the question mean? :confused:
 
  • #10
Its ok, i found out the answer they have is incorrect. Using P/V to get I gives me around 15A using the r.m.s. and around 10A using the peak value.

They are using a current of 21.2A - where that comes from i don't know.

Obviously they made a mistake.

Thanks for all the replies
 

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