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^-7T ∙ 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

Using the formula for the magnetic field due to a long straight wire, we have:

B = μ0*I/(2π*r)

where B is the magnetic field, I is the current, r is the distance from the wire, and μ0 is the permeability of free space.

Setting B equal to the strength of the Earth's magnetic field, we can solve for r:

5.0*10^-5 T = (4π*10^-7 T∙m/A)*(5.0 A)/(2π*r)

Simplifying:

r = (4π*10^-7 T∙m/A)*(5.0 A)/(2π*5.0*10^-5 T)

r = 2.0*10^-2 m = 2.0 cm

Therefore, the distance from the central axis of the wire where the magnetic field equals the strength of the Earth's magnetic field is 2.0 cm.

At What Distance from the Central Axis of a Long