The magnitude of the magnetic field (\(B\)) at a point on the axis of a circular coil carrying a current can be calculated using the formula:
\[ B = \frac{\mu_0 \cdot I \cdot N}{2 \cdot R} \]
Where:
- \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}\))
- \(I\) is the current in the coil (0.420 A)
- \(N\) is the number of turns in the coil (550 turns)
- \(R\) is the distance from the center of the coil to the point on the axis (6.00 cm converted to meters)
Let's calculate it step by step:
First, convert the distance \(R\) to meters: \(6.00 \, \text{cm} = 0.06 \, \text{m}\).
Now, plug in the values and calculate the magnetic field \(B\):
\[ B = \frac{(4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}) \cdot (0.420 \, \text{A}) \cdot (550 \, \text{turns})}{2 \cdot 0.06 \, \text{m}} \]
\[ B \approx 1.47 \times 10^{-5} \, \text{T} \]
So, the magnitude of the magnetic field at a point on the axis of the coil, 6.00 cm from its center, is approximately \(1.47 \times 10^{-5}\) Tesla.