Question

Two long, straight, parallel wires of length 3.7 m carry parallel currents of 3.6 A and 1.6 A. (a) If the wires are separated by a distance of 3.1 cm, what is the magnitude of the force between the two wires

Answer 1

Answer:

$|F|=1.37\times 10^{-4}\ N$

Explanation:

Given:

Length of the parallel wires (L) = 3.7 m

Current in the first wire (I₁) = 3.6 A

Current in the second wire (I₂) = 1.6 A

Separation between the parallel wires (d) = 3.1 cm = 0.031 m  [1 cm = 0.01 m]

Both the wires will attract each with forces equal in magnitude and opposite in direction.

Now, the magnitude of the force acting between the two wires is given as:

$|F|=\frac{\mu_0I_1I_2L}{2\pi d}$

Where, $\mu_0\to permeability\ constant=4\pi \times 10^{-7}\ N/A^2$

Plug in the given values and solve for |F|. This gives,

$|F|=\frac{(4\pi\times 10^{-7}\ N/A^2)(3.6\ A)(1.6\ A)(3.7\ m)}{2\pi\times 0.031\ m}\\\\|F|=1.37\times 10^{-4}\ N$

Therefore, the magnitude of the force between the two wires is $|F|=1.37\times 10^{-4}\ N$