Question

Four long parallel wires pass through the corners of a square with side 0.62 m. All four wires carry the same magnitude of current I = 24.1 A in the directions indicated. Find the magnetic field at point P, the midpoint of the top side of the square.

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Answer #1

ED = 0.31 m , CD = 0.62 m

VED2 + CD2-V0.312 +0.622 0.693m

0.613 12 Magnetic field by wire at B:: B, =(10-7) 2 (24.12 = 69.6 x、-r

In triangle EDC

ED = 0.31 m   , CD = 0.62

so tan\theta = CD/ ED = 0.62 /0.31

\theta = 63.4

B1 and B2 are making angle \theta with the upper side of the square at P as shown

vertical Component of B1 = B1 sin\theta   in Up

vertical Component of B2 = B2 sin\theta   in down

So The vertical components of magnetic field cancel out being in opposite direction and horizontal component adds to give total magnetic field as

Btotal = B1 Cos\theta + B2 Cos\theta

B1 = B2

so Btotal = B1 Cos\theta + B1 Cos\theta = 2 B1 Cos\theta = 2 x (69.6 x 10-7 T) Cos 63.4 = 62.33 x 10-7 T

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