A horizontal wire of length 3.0 m carries a current of 6.0 A and is oriented so that the current direction is 50o S of W. The horizontal component of the Earth's magnetic field is due north at this point and has a strength of 0.14 x 10-4 T. What is the size of the force on the wire?

Question

A horizontal wire of length 3.0 m carries a current of 6.0 A and is oriented so that the current direction is 50o S of W. The horizontal component of the Earth's magnetic field is due north at this point and has a strength of 0.14 x 10-4 T. What is the size of the force on the wire?

BlackTom AI · Accepted Answer

We start by identifying the given quantities: a horizontal wire of length L = 3.0 m carries current I = 6.0 A, with the current direction 50° South of West. The Earth’s magnetic field component in the horizontal plane points due north with magnitude B = 0.14 × 10^-4 T = 1.4 × 10^-5 T. The magnetic force on a current-carrying wire is F = I L B ⊤ sin(theta), where theta is the angle between the current direction and the magnetic field direction.

First, determine the angle between the current direction and the magnetic field:
- The current direction has components toward West and toward South, forming a vector at 50° toward the South from the West axis. In standard coordinates (x to the East, y to the North), this direction has x = -cos(50°) and y = -sin(50°).
- The magnetic field is purely northward, with components (0, +B).
- The angle between these two vectors is found from their dot product: cos(theta) = (I_hat • B_hat) = (0, +1) • (-cos50°, -sin50°) = -sin50°.
- Therefore theta = arccos(-sin50°) ≈ arccos(-0.7660) ≈ 140°. The sine of this angle, sin(theta), is sin(140°) = sin(40°) ≈ 0.6428.

Now plug in the values:
F = I L B sin(theta) = (6.0 A)(3.0 m)(1.4 × 10^-5 T)(0.6428)
= 18 × 1.4 × 10^-5 × 0.6428 N
= (25.2 × 10^-5) × 0.6428 N
≈ 1.62 × 10^-4 N.

Expressing this in the same format as the answer choices, F ≈ 1.6 × 10^-4 N. This matches the option that lists '1.6 x 10-4 N'. The other options can be ruled out by noting that the product I L B is on the order of 2.5 × 10^-4 N before accounting for the angle, and the angle reduces the force to the value near 1.6 × 10^-4 N rather than the larger or smaller magnitudes given.

To check intuition: if the current were aligned with the magnetic field, sin(theta) would be zero and the force would vanish; if perpendicular, sin(theta) would be 1 and you would get a larger force, but here the angle is neither 0 nor 90 degrees, leading to a moderate reduction from the maximum possible value.

A horizontal wire of length 3.0 m carries a current of 6.0 A and is oriented so that the current direction is 50o S of W. The horizontal component of the Earth's magnetic field is due north at this point and has a strength of 0.14 x 10-4 T. What is the size of the force on the wire?Single choice

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