S. 8/18 points Provious Anawers MI4 17.7.033. My N Electron current A wire throu

Question

S. 8/18 points Provious Anawers MI4 17.7.033. My N Electron current A wire through which a current is nowing lies along the x-axis as shown. Connecting wires which are not shown in the diagram connect the ends of the wire to batteries (which are also not shown). Electron current fiows through the wire in the -x direction, as indicated in the diagram. To calculate the magnetic field at location A due to the current in the wire, we divide the wire into pieces, approximate each piece as a point charge moving in the direction or conventional current, and calculate the magnetic field at the observation location due only to this piece; then add the contributions of all pieces to get the net magnetic field The wire is 1.4 m long, and is divided into 8 pieces. The observation location A is located at m. The conventional current running through the wire is 9.5 amperes. In this exercise you will calculate the magnetic field at the observation location due only to segment 4 of the wire. What is the direction of conventional current in this wire? How long is segment 4? 175 what is the magnitude of the vector , for segment 4? IATI = .175 what is the vector 1 for segment 4? (.175, 0,0) What is the location of the center of segment 4? source location--.0875 | × m What is the vector from source to observation location, for segment 4? What is the unit vector ? what is 7xr? Calculate the magnetic field at the observation location due only to the current in segment 4 of the wire. tesla Additional Materials

Explanation / Answer

Location of the center of segment 4: < -0.0875, 0, 0 > (The answer should have been in the vectorial form, that could be the reason it might be showing incorrect mark)

Vector r from source to location: Vector of location - Vector of source
< -0.0875, 0, 0> - <0.087, 0.192, 0> = < 0.1745 , 0.192, 0>

Unit Vector : (0.1745 i + 0.192 j + 0 k)/ ( 0.17452 + 0.1922 + 02)1/2 = 0.672 i + 0.741 j + 0 k

Cross product between two vectors a and b is a x b = |a| |b| sin
First we need to find the angle theta, that can be found by taking the dot product a.b
a.b =  |a| |b| cos
cos = 0.671
= 47.85 degrees

Now, cross product is a x b = <0.175, 0, 0> x < 0.672, 0.741, 0 > = <0, 0 , 0.1296>

You can calculate the magnetic field due to a current carrying wire element by the formula [0/4*pi][I dl sin/ r2]
We have , r(distance vector), length of the current carrying element so the magnetic field can be easily found.

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