A circular metal loop of radius 10 cm and three long straight wires carry currents of

And
As shown in the figure.Each of the straight wires is 20 cm from the center of the loop.The axes are shown in the figure,with the +z-axis coming out of the paper.What is the z component of the resultant magnetic field at the center of the loop? (μ₀ = 4π × 10^-7 T ∙ m/A)

Question

A circular metal loop of radius 10 cm and three long straight wires carry currents of

And

As shown in the figure.Each of the straight wires is 20 cm from the center of the loop.The axes are shown in the figure,with the +z-axis coming out of the paper.What is the z component of the resultant magnetic field at the center of the loop? (μ₀ = 4π × 10^-7 T ∙ m/A)

Quizplus · Accepted Answer

Using the right-hand rule, we can find the direction of the magnetic field created by each straight wire. At the center of the loop, the magnetic field created by wire 1 and wire 3 both have a positive z component, while the magnetic field created by wire 2 has a negative z component.

The magnitude of the magnetic field created by each wire can be found using the Biot-Savart law:

B1 = (μ0/4π) * (I1/0.2) = (4π × 10^-7 T ∙ m/A) * (5A/0.2m) = 0.628 T
B2 = (μ0/4π) * (I2/0.2) = (4π × 10^-7 T ∙ m/A) * (3A/0.2m) = 0.377 T
B3 = (μ0/4π) * (I3/0.2) = (4π × 10^-7 T ∙ m/A) * (4A/0.2m) = 0.502 T

The net magnetic field at the center of the loop is the vector sum of these three magnetic fields. Since the magnetic field created by wire 2 is in the opposite direction of the magnetic fields created by wires 1 and 3, we need to subtract the magnitude of B2 from the sum of the magnitudes of B1 and B3:

Bnet = sqrt(B1^2 + B2^2 + B3^2 - 2B1B2cos(120°)) = 0.36 T

The z component of this net magnetic field is negative, since the magnetic field created by wire 2 has a negative z component:

Bz = -Bnet*cos(30°) = -0.31 T = -310 μT

Therefore, the answer is A) -360 μT.

A Circular Metal Loop of Radius 10 Cm and Three