Well, well, well, looks like we've got some electrifying wires there! Let's calculate the magnetic field at the origin.
First, let's take a look at each wire separately. The top wire is carrying a current of 20 A, the middle wire is carrying 40 A, and the bottom wire is carrying 60 A.
To find the magnetic field at the origin, we can use the right-hand rule for a wire carrying current. Point your thumb in the direction of the current, and your fingers will curl in the direction of the magnetic field.
Since the middle wire is passing through the origin, the magnetic field due to this wire cancels out. So we only need to consider the top and bottom wires.
The magnetic field due to a long, straight wire at a distance r is given by the formula B = (μ₀ * I) / (2π * r), where μ₀ is the vacuum permeability.
Let's calculate the magnetic field at the origin due to the top wire carrying 20 A. Since it's at a distance of 2 m from the origin, we have:
B(top) = (4π * 10^(-7) T·m/A * 20 A) / (2π * 2 m)
After some mathy calculations, we find that B(top) = 20 × 10^(-7) T.
Next, let's calculate the magnetic field at the origin due to the bottom wire carrying 60 A. It's at a distance of 3 m from the origin, so we have:
B(bottom) = (4π * 10^(-7) T·m/A * 60 A) / (2π * 3 m)
Again, with some number crunching, we find that B(bottom) = 40 × 10^(-7) T.
Now, to find the total magnetic field at the origin, we simply add up the individual fields from the top and bottom wires:
B(total) = B(top) + B(bottom)
Oh dear, it seems I've gotten carried away with the calculations, haven't I? But fear not! I have a surprise for you. The final answer is... drumroll, please... B(total) = 60 × 10^(-7) T!
The magnitude of the magnetic field at the origin is therefore 60 × 10^(-7) Tesla. And that's how we zap our way through this magnetic field question!