# At what distance from a long straight wire carrying a current of 5.0 Amps is the magnitude of the magnetic field due to the wire equal to the strength of earth's magnetic field of about 5.0 x 10^-5...

At what distance from a long straight wire carrying a current of 5.0 Amps is the magnitude of the magnetic field due to the wire equal to the strength of earth's magnetic field of about 5.0 x 10^-5 T ?

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### 1 Answer

The magnetic field due to a current flowing through an infinitely long wire is given by `B = (mu*I)/(2*pi*r)` where B is the magnetic field, mu is the permeability of free space and equal to `4*pi*10^-7` T*m/A, and r is the radial distance.

The Earth's magnetic field is approximately `5*10^-5` T. If a wire carries a current of 5 A, the magnitude of the magnetic field due to the current is `5*10^-5` T at a distance r where:

`5*10^-5 = (4*pi*10^-7*5)/(2*pi*r)`

=> r = `(4*pi*10^-7*5)/(2*pi*5*10^-5)`

=> r = `(2*10^-2)`

**The magnetic field around the wire is `5*10^-5` T at a distance `2*10^-2` m from the wire.**

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