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A closely wound, circular coil with a diameter of 4.10 cm has 550 turns and carries a current of 0.420 A. Part B What is the magnitude of the magnetic field at a point on the axis of the coil a distance of 6.00 cm from its center?

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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.
User Jthomas
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