g Two long parallel wires are a center-to-center distance of 2.50 cm apart and carry equal anti-parallel currents of 2.70 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R

The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.

The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.

Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.

The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR

Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.

Substituting these into B' = μ₀i/πR, we have

B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)

B = 10.8/1.25 × 10⁻⁵ T

B = 8.64 × 10⁻⁵ T

B = 864 × 10⁻³ T

B = 864 mT

hamzaahmeds hamzaahmeds · Answer 2 · 2021-11-20T11:51:21+01:00

This question involves the concept of the magnetic field due to two current-carrying wires in the same direction, parallel to each other.

The magnitude of the magnetic field at the point P, which is equidistant from the wires is "8.64 x 10⁻⁵ T".

The following formula is used to find the magnetic field at the center distance between two parallel current-carrying wires in the same direction:

[tex]B = \frac{\mu_oI_1}{2\pi r}+\frac{\mu_oI_2}{2\pi r}\\\\But,\ I_1=I_2=I\\\\B = \frac{\mu_oI}{\pi r}[/tex]

where,

B = magnetic field at required point = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ H/m

I = current = 2.7 A

r = distance from wires to the point = 2.5 cm/2 = 1.25 cm = 0.0125 m

Therefore,

[tex]B=\frac{(4\pi\ x\ 10^{-7}\ H/m)(2.7\ A)}{\pi (0.0125\ m)}[/tex]

B = 8.64 x 10⁻⁵ T

Learn more about the magnetic field here:

https://brainly.com/question/23096032?referrer=searchResults