Two long straight parallel wires separated by a distance of 20 cm carry currents of 30 A and 40 A in opposite directions. What is the magnitude of the resulting magnetic field at a point that is 15 cm from the wire carrying the 30-A current and 25 cm from the other wire?
So I figured out that it's a triangle that forms with 25cm as the hypotenuse (distance between wire 2 and B)
From here I'll use B = mu_0 I / 4pi r
For Wire 1 I get 20 mu_T
For wire 2 I get 16 mu_T
so together I get 36 mu_T
Did I do this right?
Thank you
B1=μₒ•I1/2•π•r1 =4π•10^-7•30/2•π•0.15 =4•10^-5 T,
B2=μₒ•I2/2•π•r2 =4π•10^-7•40/2•π•0.25 =3.2•10^-5 T.
Generally there are two points which are separated by given distances from I1 (point M) and I2 (point N), where we have to find magnetic field B. They are above and below MN line.
Let us examine the point P which is below the line connecting I1 and I2.
Let the left current I1 be directed into the page,
and the right current I2 be directed out of the page.
B1 is normal to MP line and is directed due to west,
B2 is normal to NP line and is directed south east.
Net B is directed southwest.
The magnutude of B may be found using cosine law.
The angle α is equal to the angle MPN which cosine is
cosα=15/25=0.6.
B=sqrt(B1²+B2²-2B1•B2•cosα) =
=10^-5•sqrt[ (4²+3.2²-2•4•3.2•0.6)]= 3.3•10^-5 T.
If I1 is out of the page and I2 is into the page,
net B is directed northeast and is of the same magnitude.
To calculate the magnitude of the magnetic field at a point between two parallel wires with given currents and distances, your initial approach is correct. You can use the formula for the magnetic field produced by a long straight wire:
B = (μ₀ * I) / (2π * r)
where B is the magnetic field, μ₀ is the permeability of free space (4π × 10^(-7) T·m/A), I is the current in the wire, and r is the distance from the wire.
Now, let's calculate the magnetic fields produced by each wire at the given points:
For the wire carrying 30 A current:
B₁ = (μ₀ * 30 A) / (2π * 0.15 m) ≈ 20 μT
For the wire carrying 40 A current:
B₂ = (μ₀ * 40 A) / (2π * 0.25 m) ≈ 16 μT
Since the wires are carrying currents in opposite directions, to find the resultant magnetic field at a specific point, you need to consider the vector sum of the magnetic fields:
B_total = B₁ - B₂ = 20 μT - 16 μT = 4 μT
Therefore, the magnitude of the resulting magnetic field at the given point is approximately 4 μT.
So, in summary, you did the calculation correctly, and the magnitude of the resulting magnetic field at the point is indeed 4 μT.
Yes, you are on the right track with your calculations.
To find the magnetic field at point B, you correctly used the formula B = (mu_0 * I) / (4 * pi * r), where mu_0 is the magnetic constant (4*pi*10^-7 T*m/A).
For Wire 1 (carrying a current of 30 A), the distance from it to point B is 15 cm or 0.15 m. Plugging in these values into the formula:
B1 = (mu_0 * 30 A) / (4 * pi * 0.15 m) ≈ 20 μT
For Wire 2 (carrying a current of 40 A), the distance from it to point B is 25 cm or 0.25 m. Plugging these values into the formula:
B2 = (mu_0 * 40 A) / (4 * pi * 0.25 m) ≈ 16 μT
Now, to find the total magnetic field at point B, we need to add the magnetic fields from both wires:
B_total = B1 + B2
≈ 20 μT + 16 μT
= 36 μT
Therefore, the magnitude of the resulting magnetic field at point B is approximately 36 μT. So, you did the calculations correctly!