Calculating Magnetism on Child Homework

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The discussion centers on calculating the maximum magnetic flux density near a child sleeping 30 cm from household wiring carrying 3.6 kW of power at 240V r.m.s. Participants clarify that r.m.s. voltage represents the effective value of AC voltage, and it can be converted to peak voltage using the formula Vpeak = Vrms × √2. There is confusion over the current value used in calculations, with one participant deriving approximately 15A from the r.m.s. voltage, while the mark scheme suggests an incorrect 21.2A. The correct approach involves using the appropriate formulas for magnetic field calculations around current-carrying wires, emphasizing the distinction between average and peak values in AC circuits. Ultimately, the consensus is that the provided answer in the mark scheme is incorrect.
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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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