A current of 4.60 A is located at the origin, flowing along the z axis. (A positive current flows in the + k direction.) Find the magnetic field at the point (2.79,3.89) cm.
Enter the x and y components of the field:
Enter the x and y components of the field:
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Since your current flows along the z axis, it is effectively an infinite straight wire, and your situation has enough symmetry to make applying Ampere's Law useful:
∮ B∙ds = (μ_0) I
or
∮ B cosθ ds = (μ_0) I
Choose an Amperian Loop centered on the z axis and passing through the point (2.79,3.89) cm. You'l want to use the Pythagorean Theorem to find its radius R. With such a loop, B has a constant magnitude and so can be taken out of the integral. B is also parallel to a magnetic field line at every point around the loop, so cosθ = 1. (Symmetry considerations tell you that the magnetic field lines are circles.) So
B ∮ ds = (μ_0) I
but ∮ ds is just the circumference of the loop, 2πR.
B 2πR = (μ_0) I
Solve for the magnitude of B.
To find the components, you'll need the angle φ toward the position (2.79,3.89) cm. Note that
tanφ = 3.89 / 2.79
Once you have φ, then the x component of B will be B cosφ, and the y component will be B sinφ.
∮ B∙ds = (μ_0) I
or
∮ B cosθ ds = (μ_0) I
Choose an Amperian Loop centered on the z axis and passing through the point (2.79,3.89) cm. You'l want to use the Pythagorean Theorem to find its radius R. With such a loop, B has a constant magnitude and so can be taken out of the integral. B is also parallel to a magnetic field line at every point around the loop, so cosθ = 1. (Symmetry considerations tell you that the magnetic field lines are circles.) So
B ∮ ds = (μ_0) I
but ∮ ds is just the circumference of the loop, 2πR.
B 2πR = (μ_0) I
Solve for the magnitude of B.
To find the components, you'll need the angle φ toward the position (2.79,3.89) cm. Note that
tanφ = 3.89 / 2.79
Once you have φ, then the x component of B will be B cosφ, and the y component will be B sinφ.