Question

Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).

Answer 1

The magnetic field at point P, which is equidistant from two long parallel wires with equal anti-parallel currents, is calculated using Ampere's law. The net magnetic field is zero because the fields due to each wire cancel each other at that point.

Step-by-step explanation:

The question concerns the calculation of the magnetic field at a point equidistant from two long parallel wires that carry equal anti-parallel currents. According to the right-hand rule and Ampere's law, each wire generates a magnetic field that circles the wire. For two wires carrying currents in opposite directions, the magnetic fields at the midpoint between the wires will point in opposite directions, thus they will subtract from each other when calculating the total magnetic field at point P.

To find the magnetic field at point P, we use the formula for the magnetic field due to a long straight current-carrying wire: B = (μ₀I)/(2πd), where B is the magnetic field, μ₀ is the permeability of free space (4π x 10-7 T·m/A), I is the current, and d is the distance to the point of interest from the wire. In this case, the distance d will be the radius R = 5.00 cm since point P is equidistant from both wires.

Substituting the values into the formula, the magnetic field due to each wire at point P can be calculated. However, since the currents are anti-parallel, the net magnetic field at P would be the difference between the two fields, which is zero.

Answer 2

Image is missing, so i have attached it

Answer:

19.04 × 10⁻⁴ T in the +x direction

Step-by-step explanation:

We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.

Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)

We are given;

I = 2.4 A

R = 5 cm = 0.05 m

μ₀ is a constant = 4π × 10⁻⁷ H/m

B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)

B = 9.6 × 10⁻⁴ T

To get the direction of the field from each wire, we will use Flemings right hand rule.

From the diagram attached:

We can say the field at P from the top wire will point up/right

Also, the field at P from the bottom wire will point down/right

Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.

If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.

Let the angle be θ

Thus;

sin(θ) = (1.3/2)/5

θ = sin⁻¹(0.13) = 7.47⁰

The x component of each field would be:

9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T

Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction