At what distance from the central axis of a long straight thin wire carrying a current of 5.0 A is the magnitude of the magnetic field due to the wire equal to the strength of the Earth's magnetic field of about 5.0 × 10^-5 T? (μ₀ = 4π × 10^-7 T ∙ m/A) A) 1.0 cm B) 2.0 cm C) 3.0 cm D) 4.0 cm E) 5.0 cm

Question

Quizplus · Accepted Answer

We can use the formula for the magnetic field due to a long straight wire, given by:

B = μ₀I/2πr

Where B is the magnetic field, μ₀ is the vacuum permeability (given as μ₀ = 4π × 10^-7 T · m/A), I is the current, and r is the distance from the wire.

We want to find the distance at which the magnetic field B is equal to the strength of the Earth's magnetic field, given as B = 5.0 × 10^-5 T.

Setting these two expressions equal, we get:

5.0 × 10^-5 T = μ₀I/2πr

Solving for r, we get:

r = μ₀I/2πB

Plugging in the given values, we get:

r = (4π × 10^-7 T · m/A)(5.0 A)/(2π)(5.0 × 10^-5 T) ≈ 2.0 cm

Therefore, the answer is B) 2.0 cm.

At What Distance from the Central Axis of a Long