Find the direction and magnitude of the net magnetic field

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SUMMARY

The discussion focuses on calculating the net magnetic field generated by two parallel wires carrying currents of 5.00 A in opposite directions, separated by 10.0 cm. At the midpoint between the wires, the net magnetic field is 0 T due to equal magnitudes canceling each other. At point P1, located 10.0 cm to the right of the right wire, the net magnetic field is 5.00 x 10^-6 T directed out of the page. At point P2, 20.0 cm to the left of the left wire, the net magnetic field is 1.70 x 10^-6 T, also directed out of the page.

PREREQUISITES
  • Understanding of Ampère's Law and magnetic fields generated by current-carrying wires
  • Familiarity with the Biot-Savart Law and its application
  • Knowledge of vector addition in three-dimensional space
  • Basic proficiency in using the formula B = μI/(2πr)
NEXT STEPS
  • Study the Biot-Savart Law for calculating magnetic fields from current distributions
  • Learn about vector addition in three-dimensional magnetism
  • Explore the effects of multiple current-carrying wires on magnetic fields
  • Investigate the applications of magnetic fields in electrical engineering
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and anyone studying magnetism and electromagnetic theory, particularly those interested in the behavior of magnetic fields around current-carrying conductors.

cy19861126
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Homework Statement


The two wires shown below carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field a) at a point midway between the wires b) at point P1, 10.0 cm to the right of the wire on the right, and c) at point P2, 20.0 cm to the left of the wire on the left.

Homework Equations


B = uI/2(pi)r


The Attempt at a Solution


I got an answer for this, but I'm not sure because my instructor has never gone over vector addition in 3D nor in magnetism. Based on what I learned before, this is what I got
a) 0. because both magnetic field are at the same magnitude but opposite direction. Therefore, their magnitudes cancel and the resultant is 0

b) B = 1.26*10^-6*5.00/(2*3.14*0.1) = 1*10^-5T
B = 1.26*10^-6*5.00/(2*3.16*0.2) = 5*10^-6T
Bnet = 1*10^-5T - 5*10^-6T = 5*10^-6T out of the page
c) B = 1.26*10^-6*5.00/(2*3.14*0.2) = 5*10^-6T
B = 1.26*10^-6*5.00/(2*3.14*0.3) = 3.33*10^-6T
Bnet = 5*10^-6 - 3.3*10^-6 = 1.7*10^-6T out of page
 

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How did you get the direction of the field for part a?
 
robb_ said:
How did you get the direction of the field for part a?
There's no direction because the two direction from the two wires cancel out. One is going into the page and another one is going out of the page
 

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