Question

Two parallel wires of equal lengths are separated by a distance of 3 m from each other. The currents flowing through 1st and 2nd wire is 3 A and 4.5 A respectively in opposite directions. The resultant magnetic field at mid poin between the wires ( = permeability of free space)

Answer 1

1 x 10^{-6} T

Answer 2

Solution:

1. The magnetic field due to a long straight current-carrying wire at a distance $r$ is:

   $$B = \frac{\mu_0 I}{2\pi r}$$

   At the midpoint between the two wires (separation $= 3\, \text{m}$), the distance from each wire is:

   $$r = \frac{3}{2} = 1.5\, \text{m}$$

2. For Wire 1 (3 A):

   $$B_1 = \frac{\mu_0 \times 3}{2\pi \times 1.5} = \frac{\mu_0}{\pi}$$

3. For Wire 2 (4.5 A):

   $$B_2 = \frac{\mu_0 \times 4.5}{2\pi \times 1.5} = \frac{1.5\mu_0}{\pi}$$

4. Direction Consideration:

   Using the right-hand rule, if currents are in opposite directions, the magnetic fields at the midpoint will be in the same direction. (For example, if Wire 1 carries current upward and Wire 2 carries current downward, then at the midpoint both fields point into the page.)

5. Resultant Magnetic Field:

   $$B_{\text{net}} = B_1 + B_2 = \frac{\mu_0}{\pi} + \frac{1.5\mu_0}{\pi} = \frac{2.5\mu_0}{\pi}$$

   Substitute $\mu_0 = 4\pi \times 10^{-7}\, \text{T}\cdot\text{m/A}$:

   $$B_{\text{net}} = \frac{2.5 \times 4\pi \times 10^{-7}}{\pi} = 2.5 \times 4 \times 10^{-7} = 10 \times 10^{-7} = 10^{-6}\, \text{T}$$

Explanation (minimal core solution):

• Use $B = \mu_0 I/(2\pi r)$ with $r = 1.5\, \text{m}$.
• Compute $B_1 = \mu_0/\pi$ and $B_2 = 1.5\mu_0/\pi$.
• Adding, $B_{\text{net}} = 2.5\mu_0/\pi = 10^{-6}\, \text{T}$.