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Magnitude of magnetic field from current carrying wire?
For point 1, the magnetic field due to the horizontal wire and diagonal wire add up. The answer I get is C.
(μ₀×I) / (πd)
This is correct.
1) For point 2, I'm not sure how to find the magnitude of the magnetic field due to the diagonal wire? The answer is F apparently.
2) Why did we not have to account for the angle for point 1, but now it matters for point 2? Because the horizontal wire is in between?
1 Answer
- az_lenderLv 73 years agoFavorite Answer
For point 1, you DO account for the "angle" -- in that the PERPENDICULAR distance from point 1 to each wire is "d", so the field due to each wire is
(mu-nought)I/(2*pi*d),
and you add these two fields together to get answer C.
For point 2, the two fields due to the two wires will point in opposite directions, so you will subtract
(mu-nought)I/(2*pi*d) - (mu-nought)I/4*pi*d to get answer F.
