To find the magnetic field at the point (0, 0, 4 cm) due to a circular current-carrying loop, we can indeed use the formula for the magnetic field along the axis of a circular ring. Let's break down the problem step by step to ensure we arrive at the correct answer.
Understanding the Setup
We have a circular loop of radius \( r = 3 \) cm (since \( x^2 + y^2 = 9 \) cm² implies \( r = \sqrt{9} = 3 \) cm) carrying a current \( I = 2.5 \) A. The point where we want to calculate the magnetic field is located at (0, 0, 4 cm), which is along the z-axis, 4 cm above the center of the loop.
Magnetic Field Formula
The magnetic field \( B \) along the axis of a circular loop can be calculated using the formula:
B = \frac{{\mu_0 I r^2}}{{2 (r^2 + z^2)^{3/2}}}
Where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).
- \( I \) is the current in amperes.
- \( r \) is the radius of the loop in meters.
- \( z \) is the distance from the center of the loop to the point along the axis in meters.
Substituting Values
First, we need to convert the radius and the height from centimeters to meters:
- \( r = 3 \, \text{cm} = 0.03 \, \text{m} \)
- \( z = 4 \, \text{cm} = 0.04 \, \text{m} \)
Now, substituting these values into the formula:
B = \frac{{(4\pi \times 10^{-7}) (2.5) (0.03)^2}}{{2 ((0.03)^2 + (0.04)^2)^{3/2}}}
Calculating the Components
Let's calculate the denominator first:
- \( (0.03)^2 = 0.0009 \, \text{m}^2 \)
- \( (0.04)^2 = 0.0016 \, \text{m}^2 \)
- \( (0.03)^2 + (0.04)^2 = 0.0009 + 0.0016 = 0.0025 \, \text{m}^2 \)
- \( (0.0025)^{3/2} = (0.0025)^{1.5} = 0.000125 \, \text{m}^3 \)
Now substituting back into the formula:
B = \frac{{(4\pi \times 10^{-7}) (2.5) (0.0009)}}{{2 (0.000125)}}
Final Calculation
Calculating the numerator:
- \( 4\pi \times 10^{-7} \times 2.5 \times 0.0009 = 9 \times 10^{-10} \pi \, \text{T m} \)
Now for the denominator:
- \( 2 \times 0.000125 = 0.00025 \, \text{m}^3 \)
Putting it all together:
B = \frac{{9 \times 10^{-10} \pi}}{{0.00025}} = 3.6 \times 10^{-6} \pi \, \text{T}
Converting to Standard Form
To express this in terms of \( 10^{-7} \), we can write:
B = \frac{{3.6 \pi}}{{0.25}} \times 10^{-6} = 14.4 \pi \times 10^{-7} \, \text{T}
Thus, the magnetic field at the point (0, 0, 4 cm) is approximately \( 14.4 \pi \times 10^{-7} \, \text{T} \). If your answer differs, it might be due to a calculation error in the steps above. Double-check each calculation, particularly the conversion of units and the arithmetic involved in the formula.
