Answer :
Certainly! Let's work through solving this problem step by step to determine the magnetic field 2 meters away from a 15-meter long wire carrying a 5-A current.
### Step-by-Step Solution:
#### 1. Identify the key parameters:
- Length of the wire, [tex]\( L = 15 \)[/tex] meters
- Current flowing through the wire, [tex]\( I = 5 \)[/tex] amperes
- Distance from the wire at which we need to find the magnetic field, [tex]\( r = 2 \)[/tex] meters
#### 2. Recall the formula for the magnetic field around a long, straight current-carrying conductor:
The magnetic field [tex]\( B \)[/tex] at a distance [tex]\( r \)[/tex] from a long, straight wire carrying a current [tex]\( I \)[/tex] is given by:
[tex]\[ B = \frac{\mu_0 \cdot I}{2 \pi \cdot r} \][/tex]
where [tex]\( \mu_0 \)[/tex] is the permeability of free space (vacuum) and its value is:
[tex]\[ \mu_0 = 4 \pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \][/tex]
#### 3. Substitute the known values into the formula:
[tex]\[ B = \frac{4 \pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \times 5 \, \text{A}}{2 \pi \times 2 \, \text{m}} \][/tex]
#### 4. Simplify the expression:
Notice that the factor of [tex]\( \pi \)[/tex] in the numerator and denominator will cancel out:
[tex]\[ B = \frac{4 \times 10^{-7} \times 5}{2 \times 2} \][/tex]
Simplify the fractions:
[tex]\[ B = \frac{20 \times 10^{-7}}{4} \][/tex]
Further simplification gives:
[tex]\[ B = 5 \times 10^{-7} \, \text{T} \][/tex]
So, the magnetic field 2 meters away from the wire is:
[tex]\[ B = 5 \times 10^{-7} \, \text{T} \][/tex]
### Conclusion:
The magnetic field 2 meters away from a 15-meter long wire carrying a 5-A current is [tex]\( 5 \times 10^{-7} \)[/tex] Tesla.
### Step-by-Step Solution:
#### 1. Identify the key parameters:
- Length of the wire, [tex]\( L = 15 \)[/tex] meters
- Current flowing through the wire, [tex]\( I = 5 \)[/tex] amperes
- Distance from the wire at which we need to find the magnetic field, [tex]\( r = 2 \)[/tex] meters
#### 2. Recall the formula for the magnetic field around a long, straight current-carrying conductor:
The magnetic field [tex]\( B \)[/tex] at a distance [tex]\( r \)[/tex] from a long, straight wire carrying a current [tex]\( I \)[/tex] is given by:
[tex]\[ B = \frac{\mu_0 \cdot I}{2 \pi \cdot r} \][/tex]
where [tex]\( \mu_0 \)[/tex] is the permeability of free space (vacuum) and its value is:
[tex]\[ \mu_0 = 4 \pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \][/tex]
#### 3. Substitute the known values into the formula:
[tex]\[ B = \frac{4 \pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \times 5 \, \text{A}}{2 \pi \times 2 \, \text{m}} \][/tex]
#### 4. Simplify the expression:
Notice that the factor of [tex]\( \pi \)[/tex] in the numerator and denominator will cancel out:
[tex]\[ B = \frac{4 \times 10^{-7} \times 5}{2 \times 2} \][/tex]
Simplify the fractions:
[tex]\[ B = \frac{20 \times 10^{-7}}{4} \][/tex]
Further simplification gives:
[tex]\[ B = 5 \times 10^{-7} \, \text{T} \][/tex]
So, the magnetic field 2 meters away from the wire is:
[tex]\[ B = 5 \times 10^{-7} \, \text{T} \][/tex]
### Conclusion:
The magnetic field 2 meters away from a 15-meter long wire carrying a 5-A current is [tex]\( 5 \times 10^{-7} \)[/tex] Tesla.