The horizontal component of earth's magnetic field at a certain place is `3.0 xx 10^(-5) T` and having a direction from the geographic south to geographic north. The force per unit length on a very long straight conductor carrying, a 1.2 A in east to west direction is

Options

A. `3.0xx10^(-5)Nm^(-1)`
B. `3.2xx10^(-5)Nm^(-1)`
C. `3.6xx10^(-5)Nm^(-1)`
D. `3.8xx10^(-5)Nm^(-1)`

Explanation

To solve the problem, we need to find the force per unit length on a long straight conductor carrying a current in a magnetic field. We will use the formula for the magnetic force on a current-carrying conductor:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Horizontal component of Earth's magnetic field, \( B = 3.0 \times 10^{-5} \, \text{T} \)
- Current in the conductor, \( I = 1.2 \, \text{A} \)
- Direction of the current: East to West
- Direction of the magnetic field: From South to North

2. **Determine the Angle Between the Current and the Magnetic Field:**
- The current flows from East to West, and the magnetic field is directed from South to North. Therefore, the angle \( \theta \) between the current and the magnetic field is \( 90^\circ \).

3. **Use the Formula for Magnetic Force:**
- The force \( F \) on a current-carrying conductor in a magnetic field is given by:
\[ F = BIL \sin \theta \]
- Since we are looking for the force per unit length \( \frac{F}{L} \), we can simplify the equation:
\[ \frac{F}{L} = BI \sin \theta \]

4. **Substitute the Values:**
- We know \( \sin 90^\circ = 1 \), so:
\[ \frac{F}{L} = B \cdot I \cdot 1 \]
- Now substituting the values:
\[ \frac{F}{L} = (3.0 \times 10^{-5} \, \text{T}) \cdot (1.2 \, \text{A}) \]

5. **Calculate the Result:**
- Performing the multiplication:
\[ \frac{F}{L} = 3.6 \times 10^{-5} \, \text{N/m} \]

6. **Final Answer:**
- The force per unit length on the conductor is:
\[ \frac{F}{L} = 3.6 \times 10^{-5} \, \text{N/m} \]