We will consider here a DC circuit, made up two conductors ( go and return, or + and - ), with infinitely long, straight conductors.

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1 How to calculate the magnetic field fom a two-coe DC cable We will conside hee a DC cicuit, made up two conductos ( go and etun, o + and -, with infinitel long, staight conductos. To build up the calculation in stages, conside fist just a single cuent,. The magnetic field it poduces foms concentic cicles. So at a point (,, the magnetic field is in the diection shown as (we have assumed the cuent is coming out of the page. The size of the magnetic field is given b Ampee s law: We can calculate fom Pthagoas: So μι Β π Net we want to know the vetical and hoizontal components of : Downloaded fom the website maintained b National Gid

2 θ Fail obviousl sin( cos( (Note: with the coodinates set up the wa we have, and the cuent assumed to be positive coming out of the page, a minus sign appeas in the tem. With diffeent coodinate conventions, the minus sign might be diffeent, but the final answe would not change as long as we applied ou chosen convention consistentl. ut we can calculate sin and cos of θ: θ θ Downloaded fom the website maintained b National Gid

3 Downloaded fom the website maintained b National Gid sin( cos( Hence cos( sin( Now we have calculated the field fom a single cuent. We now need to conside the two cuents that make the cicuit togethe. We choose to conside a flat cicuit with conductos spaced d apat in the diection hee, but the pinciples appl to an geomet. Conside fist the + cuent and the field it poduces: d θ +

4 The geomet is just the same as fo the cuent but with eplaced b -d/. So we can calculate the field components in eactl the same wa, eplacing with -d/: = μ + π (( d + = μ + ( d π (( d + The - tems ae just the same but with +d/ instead of -d/, and an eta minus sign because the cuent is in the opposite diection. (f we had a vetical aa instead of a hoizontal aa, the d/ tems would appea with the tems instead of the, and if the aa was moe complicated, thee would be modifications to both the and tems. So now we have the vetical and hoizontal components fo both cuents: - + = μ + π (( + d + = μ + π (( d + = μ + ( + d π (( + d + = μ + ( d π (( d + To get the total field in each diection, we simpl add these up: = μ π (( + d + μ π (( d + = μ π ( ( + d + ( d + = μ ( d π (( d + μ ( + d π (( + d + = μ π ( d ( d + + d ( + d + We could simplif these equations futhe, but usuall we would be calculating the answes numeicall, so thee is little point. We can now do one of two things, depending on what we want to know. f we want to know the total field fom the cable, we just calculate it as oot-sum-of-squaes: Downloaded fom the website maintained b National Gid

5 total = + Often, howeve, we ae inteested in not just the field fom the cable alone, but in how it adds to the eath s DC field, so as to give the total DC field (eath plus cable at a given point. This equies us to think slightl moe in thee dimensions. geomagnetic vetical total geomagnetic field, N-S, E-W geomagnetic hoizontal cables The geomagnetic field is at an angle to the vetical. We can esolve it into two components: the vetical component, and the hoizontal component, which, b definition, is noth-south with no east-west component. Fo ou cable, a distance below the gound and off to the side, we alead know how to calculate the vetical and hoizontal components and. The total vetical field is eas: it s just the vetical component of the geomagnetic field added to the vetical field fom the cable (making sue we have a consistent sign convention as to which diection we count as a positive field. Fo the hoizontal field, it s slightl moe complicated. The cable will usuall have some abita diection, neithe noth-south no east-west. So the hoizontal field fom the cable,, is also in an abita diection. We need to esolve it into the east-west and noth-south components, using the beaing of the cable and the appopiate sine o cos function. Then we can add the noth-south component of the cable field,, N-S, to the hoizontal component of the geomagnetic field. Fo the east-west component, that comes onl fom the cable,, E-W, with no geomagnetic component. Thus we get: Vetical Hoizontal, noth-south Hoizontal, east-west geomagnetic vetical + geomagnetic hoizontal +, N-S, E-W Downloaded fom the website maintained b National Gid

6 Once we ve calculated these thee components vetical, hoizontal noth-south, and hoizontal east-west we can add them b oot-sum-of-squaes to get the total DC field. We can also calculate the diection a compass would point along. n the absence of the cable, it just points along the hoizontal component of the geomagnetic field, noth-south. n the pesence of the cable, it points along the new local hoizontal field, that is, the esultant of the hoizontal noth-south and hoizontal east-west fields: Total hoizontal east-west Total hoizontal noth-south New local compass diection cables Moe complicated situations: f ou have moe than one cicuit, wok out all the sepaate components, sum all the components fom each sepaate conducto, likewise sum all the components, then take the oot-sum-of-squaes total as befoe. f the cuents aen t infinitel long o staight, beak them into shot staight sections and calculate the components fo each of these sepaatel (ou can no longe use Ampee s law as it stands, obviousl, ou have to educe it b the sin of the angle subtended b the ends of the section at the point of inteest. Downloaded fom the website maintained b National Gid

Physics 107 TUTORIAL ASSIGNMENT #8

Physics 107 TUTORIAL ASSIGNMENT #8 Physics 07 TUTORIAL ASSIGNMENT #8 Cutnell & Johnson, 7 th edition Chapte 8: Poblems 5,, 3, 39, 76 Chapte 9: Poblems 9, 0, 4, 5, 6 Chapte 8 5 Inteactive Solution 8.5 povides a model fo solving this type