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B Field Due To a Hairpin

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A long hairpin is formed by bending an infinitely long wire, as shown. If a current of 2.8 A is set up in the wire, what is the magnitude of the magnetic field at the point a? Assume R = 11 cm.

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Answer #1

let the current flow in an anti-clockwise direction,
magnetic field due to semi-circular wire = 1/2(μI/2r). (in a direction perpendicular to the plane of ur screen)

magnetic field due to upper stright wire(half infinite wire) = 1/2(μI/2πr),(in the same direction)

magnetic field due to lower stright wire(half infinite wire) = 1/2(μI/2πr). (in the same direction)

as three are in same direction add those ans = μI/4r+μI/2πr =
=4π*10-7*2.8[2/(4*π*0.11) + 1/(4*0.11) ]

=4*π*10-7*2.8*2.272

=8.141 x10-6 T

I is the current and 'r' is the radius.

answered by: ppboy
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