Two long parallel transmission lines `40.0 cm` apart carry `25.0 A` and `75.0 A `currents. Find an locations where the net magnetic field of the two wires is zero if these currents are in
(a) the same direction (b) the opposite direction

Explanation

To solve the problem of finding locations where the net magnetic field of two long parallel transmission lines is zero, we will analyze two scenarios: (a) when the currents are in the same direction and (b) when the currents are in opposite directions.

### Given Data:
- Distance between the wires, \( d = 40.0 \, \text{cm} = 0.4 \, \text{m} \)
- Current in wire A, \( I_1 = 25.0 \, \text{A} \)
- Current in wire B, \( I_2 = 75.0 \, \text{A} \)

### (a) Currents in the Same Direction

1. **Understanding Magnetic Fields**:
- The magnetic field \( B \) due to a long straight current-carrying wire at a distance \( r \) is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
- For two wires carrying currents in the same direction, the magnetic fields will add up between the wires and cancel out outside.

2. **Setting Up the Equation**:
- Let \( x \) be the distance from wire A where the magnetic field is zero.
- The distance from wire B will then be \( 0.4 - x \).
- The magnetic field due to wire A at distance \( x \):
\[ B_A = \frac{\mu_0 I_1}{2 \pi x} \]
- The magnetic field due to wire B at distance \( 0.4 - x \):
\[ B_B = \frac{\mu_0 I_2}{2 \pi (0.4 - x)} \]
- Since both fields are in the same direction, we set them equal to each other:
\[ \frac{I_1}{x} = \frac{I_2}{0.4 - x} \]

3. **Substituting Values**:
\[ \frac{25}{x} = \frac{75}{0.4 - x} \]

4. **Cross-multiplying**:
\[ 25(0.4 - x) = 75x \]
\[ 10 - 25x = 75x \]
\[ 10 = 100x \]
\[ x = 0.1 \, \text{m} = 10 \, \text{cm} \]

### (b) Currents in Opposite Directions

1. **Understanding Magnetic Fields**:
- When the currents are in opposite directions, the magnetic fields will oppose each other in the region between the wires.

2. **Setting Up the Equation**:
- Let \( x \) be the distance from wire A where the magnetic field is zero.
- The distance from wire B will be \( 0.4 + x \) (since we are considering points outside the wires).
- The magnetic field due to wire A at distance \( x \):
\[ B_A = \frac{\mu_0 I_1}{2 \pi x} \]
- The magnetic field due to wire B at distance \( 0.4 + x \):
\[ B_B = \frac{\mu_0 I_2}{2 \pi (0.4 + x)} \]
- Set the magnitudes equal since they are in opposite directions:
\[ \frac{I_1}{x} = \frac{I_2}{0.4 + x} \]

3. **Substituting Values**:
\[ \frac{25}{x} = \frac{75}{0.4 + x} \]

4. **Cross-multiplying**:
\[ 25(0.4 + x) = 75x \]
\[ 10 + 25x = 75x \]
\[ 10 = 50x \]
\[ x = 0.2 \, \text{m} = 20 \, \text{cm} \]

### Summary of Results:
- (a) The location where the magnetic field is zero when currents are in the same direction is **10 cm from wire A**.
- (b) The locations where the magnetic field is zero when currents are in opposite directions is **20 cm from wire A**.