Two parallel wires carry currents of 20A and 40A in opposite directions. Another wire carrying a current antiparallel to 20A is placed midway between the two wires. The magnetic force on it will be

Options

A. towards 20 A
B. towards 40 A
C. zero
D. perpendicular to the plane of the currents .

Answer

To solve the problem, we need to analyze the magnetic forces acting on the wire carrying a current of 20A that is placed midway between two other wires carrying currents of 20A and 40A in opposite directions.

### Step-by-Step Solution:

1. **Identify the Configuration**:
- We have three wires:
- Wire 1 (I1) carrying 20A (let's assume it goes upwards).
- Wire 2 (I2) carrying 40A (going downwards).
- Wire 3 (I3) carrying 20A (going upwards), placed midway between Wire 1 and Wire 2.

2. **Determine the Magnetic Field due to Each Wire**:
- The magnetic field (B) created by a long straight current-carrying wire at a distance (r) from it is given by the formula:
\[ B = \frac{\mu_0 I}{2\pi r} \]
- For Wire 1 (I1 = 20A) at the position of Wire 3 (midway):
- Distance from Wire 1 to Wire 3 = d/2 (where d is the distance between Wire 1 and Wire 2).
- The direction of the magnetic field due to Wire 1 at the position of Wire 3 (using the right-hand rule) will be into the page.
- For Wire 2 (I2 = 40A) at the position of Wire 3:
- The distance is also d/2.
- The direction of the magnetic field due to Wire 2 at the position of Wire 3 will be out of the page.

3. **Calculate the Magnitudes of the Magnetic Fields**:
- Let’s denote the distance between Wire 1 and Wire 2 as d. The magnetic field at the position of Wire 3 due to Wire 1 (B1) and Wire 2 (B2) can be calculated as:
\[ B_1 = \frac{\mu_0 \cdot 20}{2\pi \cdot \frac{d}{2}} = \frac{40\mu_0}{2\pi d} \]
\[ B_2 = \frac{\mu_0 \cdot 40}{2\pi \cdot \frac{d}{2}} = \frac{80\mu_0}{2\pi d} \]

4. **Determine the Net Magnetic Field at Wire 3**:
- Since B1 is into the page and B2 is out of the page, the net magnetic field (B_net) at Wire 3 will be:
\[ B_{net} = B_2 - B_1 = \frac{80\mu_0}{2\pi d} - \frac{40\mu_0}{2\pi d} = \frac{40\mu_0}{2\pi d} \]
- The net magnetic field is directed out of the page.

5. **Calculate the Magnetic Force on Wire 3**:
- The magnetic force (F) on a current-carrying wire in a magnetic field is given by:
\[ F = I \cdot L \cdot B \cdot \sin(\theta) \]
- Here, I = 20A (current in Wire 3), L is the length of Wire 3, and θ = 90° (since the current is perpendicular to the magnetic field).
- Therefore, the force on Wire 3 (F3) is:
\[ F_3 = 20A \cdot L \cdot \frac{40\mu_0}{2\pi d} \]
- The direction of the force, using Fleming's Left-Hand Rule, will be towards Wire 2 (the 40A wire).

6. **Conclusion**:
- The net magnetic force on the wire carrying 20A (Wire 3) will be directed towards the wire carrying 40A.

### Final Answer:
The magnetic force on the wire carrying 20A will be directed towards the wire carrying 40A.