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Magnetic field direction

  • #1

Homework Statement:

Figure 29-48 shows two very long straight wires (in cross section) that each carry a current of
4.00 A directly out of the page. Distance d1 = 6.00 m and distance d2 = 4.00 m. What is the magnitude of the net magnetic field at point P, which lies on a perpendicular bisector to the wires?

Homework Equations:

B = (μ*i)/(2*π*r)
Problem1.JPG

Again struck up with the direction of the magnetic field, i suppose now the field not simple along the x axis. How to find the angle and the direction of the field. My attempt is
B1 = (μ*i)/(2*π*r) = (4*π*4)/(2*π*4) = 200nT where r = d2 = 4; is the field due to i1.
B2 = (μ*i)/(2*π*r) = (4*π*4)/(2*π*4) = 200nT where r = d2 = 4; is the field due to i2.

B1 and B2 are making an angle of 90° between them (not convinced either). The net field is √(B1^2 + B2^2 = 282 nT.
Boutput.JPG
 

Answers and Replies

  • #2
774
312
There are several issues here. First, r is not 4m. You need to do some trigonometry to find r for each wire. Second, you say the fields are perpendicular to each other. There is no reason to think that. In your drawing you show the fields are perpendicular to the two r’s. That is correct. Third, your drawing is really messed up. You show two 90 degree angles, but the lines aren’t anything like 90 degrees apart. It really matters if you are going to be able to do this correctly. Fourth, regarding direction, have you been taught the right hand rule? You know the direction of current. You should be able to determine the direction of the field. I can say that one of your directions is wrong. Finally, the fields add vectorial. Have you been taught vector addition?
 
  • #3
Ok i will redraw clearly and rework on the points you mentioned.
 
  • #4
Yes i did work on this found new tool to draw, the new diagram is
UpdatedPic.JPG


The other corrections the distance from I1 to P and I2 to P is 5m.
The B1 = B2 = (4*π *(10^-7)*4)/(2*π *5) = 160nT
The net field B = √(B1^2 + B2^2 = 226nT. Can you please advice if it is ok.
 
  • #5
774
312
Yes, much better. The distances are indeed 5m, and you now have the right hand rule directions correct. The resultant is, as you show, pointed up. Your calculation of the field strengths is also correct. You apparently have noticed that the horizontal components cancel by symmetry and you only need to calculate the vertical component (although you may want to state that)

However, your combining of the fields is incorrect. Here again I think your drawing has let you down. The Pythagorean theorem would be correct if the fields met at a right angle, but do they? What are the angles of a 3,4,5 triangle?
 
  • #6
ZapperZ
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Yes i did work on this found new tool to draw, the new diagram is
View attachment 255638

The other corrections the distance from I1 to P and I2 to P is 5m.
The B1 = B2 = (4*π *(10^-7)*4)/(2*π *5) = 160nT
The net field B = √(B1^2 + B2^2 = 226nT. Can you please advice if it is ok.
I think you have an issue with geometry. This is not a criticism, but rather I'm trying to point out the possible source of your difficulties.

The magnetic field vector must be tangential to the magnetic field lines for each current source. This may have an understanding of this, but the geometrical sketch is wrong. Figure out what is different between your sketch, and this one:

magnetic field vector.jpg

Zz.
 
  • #7
Yes finally a very tough one, the angle between B1 and B2 fields is 73.72°, B1 with X axis is 53.14° and B2 with X axis is 53.14°. The final result is
Along X-axis B1*cos(53.14) - B2*cos(53.14) = 0;
Along Y-axis B1*sin(53.14) + B2 *sin(53.14) = 256 in the +ve Y direction.
Am I correct? Please advise.
 
  • #8
Figure out what is different between your sketch, and this one:
Zz.
Yes I made a mistake in drawing the previous post as the angle calculations were wrong. Now the field B1 will not align with the line connecting the I2 with the P point and similarly for B2 field.
 
  • #9
774
312
Yes finally a very tough one, the angle between B1 and B2 fields is 73.72°, B1 with X axis is 53.14° and B2 with X axis is 53.14°. The final result is
Along X-axis B1*cos(53.14) - B2*cos(53.14) = 0;
Along Y-axis B1*sin(53.14) + B2 *sin(53.14) = 256 in the +ve Y direction.
Am I correct? Please advise.
Yes, I think you have it now. Well done. I think this is a perfect example of why step one is always making a really good really accurate really big (helps with the accuracy) diagram.

Also, regarding your sines and cosines, you didn’t actually have to figure them out. You could note the law of similar triangles to show the vertical components are 4/5 of each field.
 

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