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If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.503 X 104 T) at a distance of 15 cm from the wire, what is the maximum current the wire can carry?

User Adrielle
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1 Answer

5 votes

Answer:

37.725 A

Step-by-step explanation:

B = magnitude of the magnetic field produced by the electric wire = 0.503 x 10⁻⁴ T

r = distance from the wire where the magnetic field is noted = 15 cm = 0.15 m

i = magnitude of current flowing through the wire = ?

Magnetic field by a long wire is given as


B = (\mu _(o))/(4\pi )(2i)/(r)

Inserting the values


0.503* 10^(-4) = (10^(-7))(2i)/(0.15)

i = 37.725 A

User Kannan Kandasamy
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