A straight copper wire that is 1 milimeter in diameter carries a current of 20 miliamps. Whas the magnitude of the largest magnetic field created by this wire in Tesla?
To determine the magnitude of the largest magnetic field created by the wire, we can use the Biot-Savart law, which relates the magnetic field created by a current-carrying wire to the current, distance, and geometry of the wire.
The Biot-Savart law formula is given by:
B = (μ₀ / 4π) * (I * dl × r) / r²
Where:
- B is the magnitude of the magnetic field
- μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A)
- I is the current in Amperes
- dl is an infinitesimally small section of the wire
- r is the distance from the wire to the point of interest
In this case, we are interested in finding the maximum magnetic field at the surface of the wire. By applying the Biot-Savart law, we can calculate it.
The current is given as 20 milliamps, which can be converted to Amperes by dividing by 1000:
I = 20 mA / 1000 = 0.02 A
The diameter of the wire is given as 1 millimeter, and since the wire is straight, the radius is half the diameter:
r = 1 mm / 2 = 0.5 mm = 0.0005 m
Now, let's calculate the magnetic field using the Biot-Savart law:
B = ( (4π × 10⁻⁷ T·m/A) / (4π) ) * (0.02 A * 2π * 0.0005 m) / (0.0005 m)²
B = (10⁻⁷) * (0.02 * 2π) / (0.0005)
B ≈ 1.257 * 10⁻⁶ T
Therefore, the magnitude of the largest magnetic field created by the wire is approximately 1.257 μT (microtesla).