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QUESTION 1 at the location another vrire that is 8.4 cm away: How much current flowing wice creales magnetic field of magnitude 000057 the wire creating this fleld?...

Question

QUESTION 1 at the location another vrire that is 8.4 cm away: How much current flowing wice creales magnetic field of magnitude 000057 the wire creating this fleld?

QUESTION 1 at the location another vrire that is 8.4 cm away: How much current flowing wice creales magnetic field of magnitude 000057 the wire creating this fleld?



Answers

SSM In Fig. 29-44, point $P_{2}$ is at perpendicular distance $R=$ $25.1 \mathrm{~cm}$ from one end of a straight wire of length $L=13.6 \mathrm{~cm}$ carrying current $i=0.693$ A. (Note that the wire is not long.) What is the magnitude of the magnetic field at $P_{2}$ ?

So in this problem, we're given the specifications for a solid oId. And so, given a applied current to its we have a magnetic field inside of it, and we want to figure out then what it's permeability must be. So we can just say we know the magnetic field inside of Stalin. OId is just mu. So the permeability of us Illinois, which we want to find times and the number of turns of coil been divided by L the length of the cell annoyed and times I the current. Ah, lot of times we writes this using little and equals and over l. So this would be the density of the turns were not given that here directly. So instead Welles will use and the total number of turns divided by all the length. Ah, So from this equation, we can find that the permeability then would just be the magnetic field inside times l over end times I and so we have all these values. So we have to point to Tesla for the field and the length of the Stalin. OId is 0.0.38 meters and we have 640 turns and we're applying occurrence of 48 amps. So we plug all that in and we should get that Mu is 2.7 times 10 to the minus fifth. And that's in units of Tesla's times, meters over amps. And so, if we divide this, then my mu knots, we're going to see that it's about 22 times bigger. So this is about 22 times you not.

In this example. We're trying to find the current in a wire when there is a magnetic field, a certain radius away. So first thing is, we know that our magnetic food field equation for a single long wear is equal to you, not I over two pi. Okay, Where a is the radius or distance away from the wire. So since we're trying to find the current, we can just rearrange this. So before I So I is equal. Teoh b times two pi a all over Munich and we can just plugging these values we get current equal to the magnetic field is 67 micro Tesla So 67 I'm standing with minus six times two pi times are radius value of 1.2 centimeters. And by this all by the constant mu not just 1.26 times 10 to the minus six. We get a current to exactly four amps

In order to solve the given problem, we take X axis XX is to be along the wire with the origin it there right midpoint the current in opposed to ex direction, direction All the segments off the wire producing Mandy feel that point B two that are out off the page. According to the bio, say what laws? The magnitude of fielded in fine at the symbol Sigmund policed at the P two is going by Deke the physical to mu nod I time sign off if it are divided by four by all square downs of the ex where that is the angle between the segment and line drawn from the segment to pee to our is the length off that line which are the functions off ex replacing we are is equal to my x square plus are square and sign theta maior or kept a lot of where our, which is already riding by X squared plus are square we can write the total field b will be called to you know I are divided by four by integral minus l 20 d x divided by x square plus are square This is kept a large square, three over two then taking integration and putting limits. We get mu nod I divided by four by our l divided by square root off l squared Plus are square Next step we substitute the values and the great to feel we get here is off the substitution 1.32 dimes Dental power minus seven Tesler.

For the given problem. Our ex axes is along the while with original meet point. The current flows in opposed to direction poster, ex direction All the segments, all for wild, produced a Manti Feel that one. They're done out off the page. According to the Bio Support's law, the majored off Field it in finite. A single segment produces a TP. One is given by D. B is a call to new I you know times I divide by four by times a sign later. Derive our square DX where third eye's dangle between segment and the line drawn from the segment to be one and are the lent off deadline are the functions off ex Replacing our with x square plus are square and sine theta We the are or our that is our derive a square root off X squared plus are square We integrate from X is equal to minus l over too to l A X is equal to Ellen Would do the total field is then given by B is equal to north times I r you're I buy for boy integration from minus l want to do l will do the X divided by X square, plus all square. This is through your two, which is equal Thio. We're in the values off limits, you know? And I do. I buy to buy our l Divide by school route off l squared plus four are square The next step we substitute the values some student values we get The B B to B B is equal to 5.3 times dental power minus eight Tesla.


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