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If an electric wire is allowed to produce a magnetic field no larger than that of earth...
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The magnetic field of a long straight wire at the distance d away from the wire is determined by the formula
where I is the current in the wire and mu_0 is the magnetic permeability constant
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
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.
Posted by ishpiro on April 10, 2013 at 1:05 AM (Answer #1)
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
Posted by justaguide on April 10, 2013 at 3:30 AM (Answer #2)
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