Two long straight wires P and Q carring equal current 10 A each were kept parallel to each other at 5 cm distance.Mangnitude of magnetic force experienced by 10 cm length of wire p is F_1- if distance between wires is halved and currents on them are doubled , force F_2 on 10 cm length of wire P will be:

Options

A. `10F_1`
B. `8F_1`
C. `F_(1)/8`
D. `F_(1)/10`

Answer

To solve the problem, we need to determine the magnetic force experienced by wire P when the distance between the wires is halved and the currents are doubled. We can use the formula for the magnetic force between two parallel current-carrying wires.

### Step-by-Step Solution:

1. **Understand the Formula**: The magnetic force \( F \) per unit length between two long parallel wires carrying currents \( I_1 \) and \( I_2 \) separated by a distance \( d \) is given by:
\[ F = \frac{\mu_0 I_1 I_2}{2\pi d} L \]
where \( \mu_0 \) is the permeability of free space, \( L \) is the length of the wire, and \( d \) is the distance between the wires.

2. **Calculate Initial Force \( F_1 \)**: For the initial conditions:
- Currents \( I_1 = I_2 = 10 \, \text{A} \)
- Distance \( d = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Length \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)

Substituting these values into the formula:
\[ F_1 = \frac{\mu_0 (10)(10)}{2\pi (0.05)} (0.1) \]

3. **Simplify \( F_1 \)**:
\[ F_1 = \frac{\mu_0 \cdot 100}{2\pi \cdot 0.05} \cdot 0.1 \]
\[ = \frac{\mu_0 \cdot 100 \cdot 0.1}{2\pi \cdot 0.05} \]
\[ = \frac{10 \mu_0}{\pi} \]

4. **Calculate New Force \( F_2 \)**: Now consider the new conditions:
- Currents are doubled: \( I_1 = I_2 = 20 \, \text{A} \)
- Distance is halved: \( d = 2.5 \, \text{cm} = 0.025 \, \text{m} \)

Substitute these new values into the formula for \( F_2 \):
\[ F_2 = \frac{\mu_0 (20)(20)}{2\pi (0.025)} (0.1) \]

5. **Simplify \( F_2 \)**:
\[ F_2 = \frac{\mu_0 \cdot 400}{2\pi \cdot 0.025} \cdot 0.1 \]
\[ = \frac{\mu_0 \cdot 400 \cdot 0.1}{2\pi \cdot 0.025} \]
\[ = \frac{40 \mu_0}{\pi} \]

6. **Find the Ratio \( F_2/F_1 \)**:
\[ \frac{F_2}{F_1} = \frac{\frac{40 \mu_0}{\pi}}{\frac{10 \mu_0}{\pi}} = \frac{40}{10} = 4 \]

7. **Final Calculation**: Since the force has increased by a factor of 4 due to the doubling of the current and halving of the distance, we can summarize:
\[ F_2 = 4 F_1 \]

### Conclusion:
Thus, if \( F_1 \) is the initial force, the new force \( F_2 \) is:
\[ F_2 = 4 F_1 \]