# If an electric wire is allowed to produce a magnetic field no larger than that of earth (5.5 x 10^-5 T) at a distance of 25 cm, what is the maximum current the wire can carry?

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### 2 Answers

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 `5.5 x 10^-5` T. If the maximum magnetic field at a distance 25 cm from the wire cannot be greater than `5.5*10^-5` , the current that it can carry is X, where:

`5.5*10^-5 = (4*pi*10^-7*X)/(2*pi*0.25)`

=> X = `(5.5*10^-5*2*pi*0.25)/(4*pi*10^-7)`

=> X = `10^2*0.6875`

=> X = 68.75 A

**The wire can carry a maximum current of 68.75 A if the magnetic field at a distance 25 cm from the wire cannot exceed `5.5*10^-5` T**

**Sources:**

The magnetic field of a long straight wire at the distance d away from the wire is determined by the formula

`B=mu_0I/(2pid)`

where I is the current in the wire and mu_0 is the magnetic permeability constant

`mu_0=4pi*10^(-7) N/A^2`

Plugging in the field of the Earth B = 5.5*10^-5 T and the distance in meters (25 cm = 0.25 m), you get

`5.5*10^(-5)=(4pi*10^(-7)I)/(2pi*0.25)`

Pi will cancel and solving for I you can find

`I=(2*5.5*10^(-5)*0.25)/(4*10^(-7))=68.75` Amps - is the maximum current the wire can carry in order not to exceed the field of the Earth at the distance 25 cm away.

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