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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?

Respuesta :

rokettojanpu

The magnetic field at a given distance from a very long (essentially infinitely long) current-carrying wire is given by:

B = μ₀I/(2πr)

B = magnetic field strength, μ₀ = magnetic constant, I = current, r = distance from wire

Given values:

B = 5.0×10⁻⁵T, μ₀ = 4π×10⁻⁷H/m, I = 5.0A

Plug in and solve for r:

5.0×10⁻⁵ = 4π×10⁻⁷(5.0)/(2πr)

r = 0.02m

okpalawalter8

We have that the distance is given as

r=0.02m

From the question we are told

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?

Generally the equation for the magnetic field strength is mathematically given as

[tex]\\ B=\frac{\mu l}{2 \pi r}[/tex]

Therefore

[tex]r=\frac{\mu l}{B*2\pi}\\\\r=\frac{4*\pi *10^(-7}*5}{5*10^{-5} *2 \pi}\\\\r=0.02m[/tex]

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