A telephonic cable at a place has four long straight horizontal wires carrying a current of `1.0amp.` in the same direction east to west. The earth's magnetic field at the place is `0.39G` and the angle of dip is `35^@`. The magnetic declination is almost zero. What are the resultant magnetic fields at points `4.0cm` below and above the cable?

Options

A. 0.25G
B. 0.50G
C. 1.25G
D. 2.50G

Explanation

To find the resultant magnetic fields at points 4.0 cm below and above the telephonic cable carrying a current, we will follow these steps:

### Step 1: Determine the Horizontal Component of the Earth's Magnetic Field (BH)
The horizontal component of the Earth's magnetic field can be calculated using the formula:
\[ BH = B \cdot \cos(\delta) \]
where:
- \( B = 0.39 \, \text{G} \) (the total magnetic field),
- \( \delta = 35^\circ \) (the angle of dip).

Calculating \( BH \):
\[ BH = 0.39 \cdot \cos(35^\circ) \approx 0.39 \cdot 0.8192 \approx 0.319 \, \text{G} \]

### Step 2: Determine the Vertical Component of the Earth's Magnetic Field (BV)
The vertical component can be calculated using:
\[ BV = B \cdot \sin(\delta) \]

Calculating \( BV \):
\[ BV = 0.39 \cdot \sin(35^\circ) \approx 0.39 \cdot 0.5736 \approx 0.224 \, \text{G} \]

### Step 3: Calculate the Magnetic Field Due to the Wires (BY)
For a long straight wire carrying current, the magnetic field at a distance \( r \) from the wire is given by:
\[ BY = \frac{\mu_0 I}{2\pi r} \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( I = 1.0 \, \text{A} \),
- \( r = 0.04 \, \text{m} \) (4.0 cm).

Calculating \( BY \):
\[ BY = \frac{(4\pi \times 10^{-7}) \cdot 1}{2\pi \cdot 0.04} = \frac{2 \times 10^{-7}}{0.04} = 5 \times 10^{-6} \, \text{T} = 0.05 \, \text{G} \]

### Step 4: Calculate the Resultant Magnetic Field Below the Cable
At a point 4.0 cm below the cable, the magnetic field due to the wires (BY) will be directed downwards (opposite to the vertical component of the Earth's field). Therefore, we can write the resultant magnetic field \( B_{below} \) as:
\[ B_{below} = \sqrt{(BH)^2 + (BV - BY)^2} \]

Substituting the values:
\[ B_{below} = \sqrt{(0.319)^2 + (0.224 - 0.05)^2} \]
Calculating:
\[ B_{below} = \sqrt{(0.319)^2 + (0.174)^2} \approx \sqrt{0.101761 + 0.030276} \approx \sqrt{0.132037} \approx 0.363 \, \text{G} \]

### Step 5: Calculate the Resultant Magnetic Field Above the Cable
At a point 4.0 cm above the cable, the magnetic field due to the wires (BY) will be directed upwards (in the same direction as the vertical component of the Earth's field). Therefore, we can write the resultant magnetic field \( B_{above} \) as:
\[ B_{above} = \sqrt{(BH)^2 + (BV + BY)^2} \]

Substituting the values:
\[ B_{above} = \sqrt{(0.319)^2 + (0.224 + 0.05)^2} \]
Calculating:
\[ B_{above} = \sqrt{(0.319)^2 + (0.274)^2} \approx \sqrt{0.101761 + 0.075076} \approx \sqrt{0.176837} \approx 0.420 \, \text{G} \]

### Final Results
- The resultant magnetic field 4.0 cm below the cable is approximately **0.363 G**.
- The resultant magnetic field 4.0 cm above the cable is approximately **0.420 G**.