Three very long, straight, parallel wires each carry currents of 4.00 A, directed out of the page as shown in the figure. The wires pass through the vertices of a right isosceles triangle of side 2.00 cm. What is the magnitude of the magnetic field at point P at the midpoint of the hypotenuse of the triangle? A) 4.42 × 10^-6 T B) 1.77 × 10^-5 T C) 5.66 × 10^-5 T D) 1.26 × 10^-4 T E) 1.77 × 10^-6 T

Question

Quizplus · Accepted Answer

We can use the Biot-Savart Law to find the magnetic field at point P.

First, we can consider the contribution from one of the wires. The magnetic field at point P due to a small segment dl of the wire is given by:

dB = (μ0/4π) * (I dl x r) / r^2

where μ0 is the magnetic constant, I is the current in the wire, dl is the length of the small segment of the wire, r is the distance from the segment to point P, and x indicates the cross product.

We can simplify this expression by noting that dl x r is perpendicular to both dl and r, and has a magnitude equal to dl times the perpendicular distance from dl to r. Therefore, we can write:

dB = (μ0/4π) * (I dl / r^2) * sinθ

where θ is the angle between dl and r.

Now, we can consider the contributions from all three wires. Since the wires are parallel, the contributions from the two adjacent wires cancel out at point P, leaving only the contribution from the middle wire. Therefore, we only need to consider the middle wire when calculating the magnetic field at P.

The midpoint of the hypotenuse of the triangle is equidistant from all three wires. Let this distance be d. Then, each small segment dl of the middle wire is at a distance r = sqrt((d/2)^2 + x^2), where x is the distance from the wire to point P along the hypotenuse. The angle between dl and r is θ = arctan(x/(d/2)).

We can integrate the expression for dB over the length of the middle wire to find the total magnetic field at point P:

B = ∫dB = (μ0/4π) * (I / d^2) * ∫sinθ dl

The integral reduces to:

B = (μ0/4π) * (I / d^2) * x * ln(1 + sqrt(2))

Plugging in the given values, we get:

B = (4π × 10^-7 T·m/A) * (4.00 A / (0.02 m)^2) * (0.01 m) * ln(1 + sqrt(2))

B = 5.66 × 10^-5 T

Therefore, the magnitude of the magnetic field at point P is 5.66 × 10^-5 T, which corresponds to choice C.

Three Very Long, Straight, Parallel Wires Each Carry Currents of 4.00