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Two long parallel wires are 30 cm apart ...

Two long parallel wires are 30 cm apart carrying currents 10 A and 15 A respectively in the same direction. The force acting over a length of 5 m of the wires is

A

`5xx10^(-4)N," (attraction)"`

B

`1xx10^(-4)N," (attraction)"`

C

`5xx10^(-4)N," (repulsion)"`

D

`1xx10^(-4)N," (repulsion)"`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the force acting on two long parallel wires carrying currents in the same direction, we will follow these steps: ### Step 1: Understand the Concept Two long parallel wires carrying currents in the same direction will exert an attractive force on each other. The formula for the force per unit length between two parallel wires is given by: \[ F/L = \frac{\mu_0 I_1 I_2}{2\pi R} \] where: - \( F \) is the force, - \( L \) is the length of the wire, - \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)), - \( I_1 \) and \( I_2 \) are the currents in the wires, - \( R \) is the distance between the wires. ### Step 2: Identify the Given Values From the problem, we have: - Current in wire 1, \( I_1 = 10 \, \text{A} \) - Current in wire 2, \( I_2 = 15 \, \text{A} \) - Distance between the wires, \( R = 30 \, \text{cm} = 0.3 \, \text{m} \) - Length of the wires considered, \( L = 5 \, \text{m} \) ### Step 3: Calculate the Force per Unit Length Using the formula for force per unit length: \[ F/L = \frac{\mu_0 I_1 I_2}{2\pi R} \] Substituting the known values: \[ F/L = \frac{(4\pi \times 10^{-7}) \times (10) \times (15)}{2\pi \times 0.3} \] ### Step 4: Simplify the Equation The \( \pi \) terms cancel out: \[ F/L = \frac{(4 \times 10^{-7}) \times (10) \times (15)}{2 \times 0.3} \] Calculating the numerator: \[ F/L = \frac{(4 \times 10^{-7}) \times 150}{0.6} \] Calculating \( 4 \times 150 = 600 \): \[ F/L = \frac{600 \times 10^{-7}}{0.6} \] This simplifies to: \[ F/L = 1000 \times 10^{-7} = 10^{-4} \, \text{N/m} \] ### Step 5: Calculate the Total Force Over the Length of 5 m Now, to find the total force acting over a length of 5 m: \[ F = (F/L) \times L = (10^{-4} \, \text{N/m}) \times (5 \, \text{m}) = 5 \times 10^{-4} \, \text{N} \] ### Step 6: Determine the Nature of the Force Since the currents are in the same direction, the force is attractive. ### Final Answer The force acting over a length of 5 m of the wires is: \[ \boxed{5 \times 10^{-4} \, \text{N}} \text{ (attractive force)} \]
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