Two long, parallel wires carry current in the x-y plane. One wire carries 30 A to the left along the x-axis. The other carries 50 A to the right along a parallel line at y = 0.28 m. At what y-axis position in meters is the magnetic field equal to zero?
To determine the y-axis position where the magnetic field is equal to zero, we can use the Biot-Savart law, which gives the magnetic field produced by a current-carrying wire.
The Biot-Savart law states that the magnetic field (B) produced at a point due to a current-carrying wire is directly proportional to the current (I), the length of the wire (dl), and the sine of the angle (θ) between the wire and the line connecting the wire to the point. Mathematically, it can be written as:
dB = (μ₀ / 4π) * (I dl * sin(θ)) / r²
Where:
- dB is the tiny magnetic field produced by a small segment of the wire
- μ₀ is the permeability of vacuum (4π x 10^-7 Tm/A)
- I is the current flowing through the wire in Amperes
- dl is the length element of the wire
- θ is the angle between the wire and the line connecting the wire to the point
- r is the distance between the wire and the point
To find the y-axis position where the magnetic field is zero, we need to find the point where the magnetic fields due to both wires cancel each other out.
Let's consider a point on the y-axis at position (0, y) and calculate the magnetic fields due to both wires at this point:
For the wire carrying 30 A to the left along the x-axis, the magnetic field at point (0, y) can be calculated using the Biot-Savart law by integrating over the entire length of the wire. Here:
- I = 30 A
- dl = dx (infinitesimal length element along the wire)
- θ = 90 degrees (since the wire is along the x-axis, perpendicular to the line connecting the wire to the point)
- r = √(x² + y²) (the distance between the wire and the point)
For the wire carrying 50 A to the right along y = 0.28 m, the magnetic field at point (0, y) can also be calculated using the Biot-Savart law. Here:
- I = -50 A (the current is negative since it is flowing in the opposite direction)
- dl = dy (infinitesimal length element along the wire)
- θ = 90 degrees (since the wire is parallel to the line connecting the wire to the point)
- r = |y - 0.28| (the distance between the wire and the point)
We will integrate these magnetic fields and equate them to find the y-axis position where they cancel out.
∫[(μ₀ / 4π) * (30 dx * sin(90)) / r²] = ∫[(μ₀ / 4π) * (50 dy * sin(90)) / |y - 0.28|²]
Simplifying the equation:
∫[30 * dx / r²] = ∫[-50 * dy / (y - 0.28)²]
Now, the left-hand side (LHS) only depends on x, while the right-hand side (RHS) only depends on y. Therefore, both sides of the equation must be equal to a constant value, which we'll call C.
∫[30 * dx / r²] = C
∫[-50 * dy / (y - 0.28)²] = C
Now, we can solve these two separate integrals:
∫dx / r² = C/30
∫-50 * dy / (y - 0.28)² = C
Integrating the left-hand side (LHS) with respect to x:
∫dx / r² = ∫dx / (√(x² + y²))² = ∫dx / (x² + y²) = arctan(y/x) = C/30
Integrating the right-hand side (RHS) with respect to y:
-50 * ∫1 / (y - 0.28)² dy = -50 * [-1 / (y - 0.28)] = 50/(0.28 - y) = C
Setting the two equations equal to each other:
arctan(y/x) = 50/(0.28 - y)
Now, we can solve this equation for y to find the y-axis position where the magnetic field is zero. However, the exact solution will involve complex calculations.
Alternatively, you can plot the two equations on a graphing software using different values of y and find the value of y where the two curves intersect or cancel each other out, giving the y-axis position where the magnetic field is equal to zero.