A copper wire of diameter 1.6 mm carries a current of 20A. Find the maximum magnitude of the magnetic field `(vec B)` due to this current.

Answer

To solve the problem of finding the maximum magnitude of the magnetic field (B) due to a current-carrying copper wire, we can follow these steps:

### Step 1: Identify the formula for the magnetic field around a straight conductor
The magnetic field (B) at a distance \( r \) from a long straight conductor carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
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 distance from the center of the wire in meters.

### Step 2: Determine the radius of the wire
Given that the diameter of the wire is \( 1.6 \, \text{mm} \), we can calculate the radius \( r \):
\[ r = \frac{\text{diameter}}{2} = \frac{1.6 \, \text{mm}}{2} = 0.8 \, \text{mm} = 0.8 \times 10^{-3} \, \text{m} = 8 \times 10^{-4} \, \text{m} \]

### Step 3: Substitute the values into the magnetic field formula
To find the maximum magnetic field, we will substitute \( I = 20 \, \text{A} \) and \( r = 8 \times 10^{-4} \, \text{m} \) into the formula:
\[ B_{\text{max}} = \frac{\mu_0 I}{2 \pi r} \]
Substituting the values:
\[ B_{\text{max}} = \frac{(4\pi \times 10^{-7}) \times 20}{2 \pi \times (8 \times 10^{-4})} \]

### Step 4: Simplify the expression
We can simplify the expression:
\[ B_{\text{max}} = \frac{(4 \times 10^{-7}) \times 20}{2 \times (8 \times 10^{-4})} \]
\[ = \frac{80 \times 10^{-7}}{16 \times 10^{-4}} = \frac{80}{16} \times 10^{-3} = 5 \times 10^{-3} \, \text{T} \]

### Step 5: Convert to milliTesla
Since \( 1 \, \text{T} = 1000 \, \text{mT} \):
\[ B_{\text{max}} = 5 \, \text{mT} \]

### Final Answer
The maximum magnitude of the magnetic field due to the current in the copper wire is:
\[ B_{\text{max}} = 5 \, \text{mT} \]