Two parallel wires separated by 16 cm are carrying currents of 14 A (left) and 20A (right) in the same direction (up the page). Determine the magnetic field at position 3 cm to the right of the 20A wire. Calculate the distance where the magnetic field would be zero.
To determine the magnetic field at a specific point due to a current-carrying wire, you can use the Biot-Savart Law. The law states that the magnetic field (B) at a point is directly proportional to the current (I), length of the wire segment (dl), and inversely proportional to the square of the distance (r) between the wire segment and the point. The equation for the magnetic field due to a current-carrying wire is:
dB = (μ0 / 4π) * (I * dl × r) / r^3,
where dB is the magnetic field due to a small wire segment dl, μ0 is the permeability of free space (approximately 4π × 10^-7 Tm/A), I is the current, r is the distance between the wire segment and the position of interest, and r^3 is the cube of the distance.
To find the magnetic field at a point 3 cm to the right of the 20A wire, we will consider an infinitesimally small wire segment on the 20A wire at a distance y from the point of interest. We will integrate the contributions of all the wire segments on the 20A wire to find the total magnetic field at the point. Let's calculate it step by step:
1. Determine the magnetic field due to a small wire segment:
- dB = (μ0 / 4π) * (I * dl × r) / r^3,
- dB = (μ0 / 4π) * (20A * dl) / r^2, since r = 3 cm = 0.03 m.
2. Integrate the contributions from all the wire segments on the 20A wire:
- B = ∫(μ0 / 4π) * (20A * dl) / r^2,
- B = (μ0 / 4π) * 20A * ∫(dl / r^2).
3. Determine the total distance the wire segments on the 20A wire cover:
- Distance = 16 cm = 0.16 m.
4. Substitute the expression dl = dx, where dx represents the infinitesimally small segment of wire on the 20A wire:
- B = (μ0 / 4π) * 20A * ∫(dx / r^2).
5. The limits of integration for x lie between -0.16 and 0, as we are integrating from the left end to the right end of the wire:
- B = (μ0 / 4π) * 20A * ∫(dx / r^2) from -0.16 to 0.
6. Integrate the expression with respect to x, to obtain the final magnetic field:
- B = (μ0 / 4π) * 20A * ∫(1 / r^2)dx from -0.16 to 0.
7. Evaluate the integral, considering the limits of integration:
- B = (μ0 / 4π) * 20A * [(1 / r^2) * (x)] from -0.16 to 0.
- B = (μ0 / 4π) * 20A * (1 / r^2) * (0 - (-0.16)).
- B = (μ0 / 4π) * 20A * (0.16 / r^2).
Now, to calculate the distance where the magnetic field would be zero, we need to determine the position at which the magnetic fields due to the 14A and 20A wires cancel each other out.
Since the currents in both wires are in the same direction, the magnetic fields due to the wires will either add up or subtract depending on the position. The magnetic field due to the 14A wire is in the opposite direction to the magnetic field due to the 20A wire.
To find the distance where the magnetic field is zero, we can set the two magnetic fields equal to each other and solve for r:
(μ0 / 4π) * 14A * (0.16 / (16cm)^2) = (μ0 / 4π) * 20A * (0.16 / r^2).
By cancelling out the common terms, we get:
14 * (0.16 / (16cm)^2) = 20 * (0.16 / r^2).
Simplifying further, we have:
14 * 0.01 = 20 * (0.16 / r^2).
Dividing both sides by 20, we get:
14 * 0.01 / 20 = 0.16 / r^2.
Simplifying, we have:
0.007 = 0.16 / r^2.
Cross multiplying, we get:
r^2 = 0.16 / 0.007.
r^2 = 22.857.
Taking the square root, we find:
r ≈ 4.78 cm.
Therefore, the distance where the magnetic field would be zero is approximately 4.78 cm.