Question

If an electric wire is allowed to produce a magnetic field no larger than that of...

If an electric wire is allowed to produce a magnetic field no larger than that of the Earth(0.50 *10^-4T at a distance of 21cm , what is the maximum current the wire can carry?
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Answer #1
Concepts and reason

The concepts used to solve this problem are magnetic field due to a current-carrying wire.

Use the relation between magnetic field due to a current-carrying wire, radial distance, and permeability of free space to find the maximum current in the wire.

Fundamentals

The Ampere’s law states that “for any closed loop, the sum of the length elements multiples the magnetic field in the direction of the length element and is equal to the permeability times the electric current enclosed in the closed loop path.

Expression for the magnetic field due to current carrying wire is

B=

Here, the magnetic field is , the radial distance is , the permeability of free space is , and the current is .

Expression for the magnetic field due to a current-carrying wire is

B=

Rearrange the above equation to get

1-B(27r)
HO

Expression for the maximum current in the wire is

Substitute 0.50x104T
for , for , and 47x10-T-m/A
for .

(0.50x10^T)(2)(a)(21cm)
1cm
I=-
41 x10-T.m/A
= 52.5 A

Ans:

The maximum current in the wire is 52.5A
.

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