A long solenoid ( radius = 1.4 cm) has a current of a 0.34 A in its winding. A long wire carrying a current of 18 A is parallel to and 1.2 cm from the axis of the solenoid. What is the magnitude of the resulting magnetic field at a point on the axis of the solenoid?
(Hint: What are the directions of the B fields created by the solenoid and the wire?)
B fields are perpendicular.
So figure each, then add as vectors.
To find the magnitude of the resulting magnetic field at a point on the axis of the solenoid, we need to consider the magnetic fields created by both the solenoid and the wire, and then calculate the resultant magnetic field.
The magnetic field created by the long solenoid can be determined using the formula:
B_solenoid = μ₀ * n * I
where B_solenoid is the magnetic field, μ₀ is the permeability of free space (4π × 10^-7 T·m/A), n is the number of turns per unit length of the solenoid, and I is the current in the solenoid.
Given that the radius of the solenoid is 1.4 cm, we can calculate the number of turns per unit length (n) using the formula:
n = N / L,
where N is the total number of turns in the solenoid and L is the length of the solenoid.
Assuming the solenoid is very long compared to its radius, we can consider it to be infinite in length. Therefore, the length of the solenoid is irrelevant in this case.
Now, we need to calculate the number of turns (N) in the solenoid. The number of turns in a solenoid can be calculated using the formula:
N = (μ₀ * μᵣ * A * n) / I,
where μᵣ is the relative permeability of the material inside the solenoid, and A is the cross-sectional area of the solenoid.
Since the relative permeability (μᵣ) is not given, we can assume it to be equal to 1 for air or vacuum.
To calculate the cross-sectional area (A) of the solenoid, we can use the formula:
A = π * r²,
where r is the radius of the solenoid.
Now, we can substitute the values into the formula for N:
N = (μ₀ * A * n) / I.
Once we have the value of N, we can calculate the value of n:
n = N / L.
Now, we have the value of n, we can substitute the values into the formula for the magnetic field:
B_solenoid = μ₀ * n * I.
We also need to consider the magnetic field created by the wire, which can be determined using the formula:
B_wire = (μ₀ * I_wire) / (2π * r_wire),
where B_wire is the magnetic field, I_wire is the current in the wire, and r_wire is the distance between the wire and the point on the axis of the solenoid.
Given that the current in the wire is 18 A, and the distance between the wire and the axis of the solenoid is 1.2 cm, we can substitute these values into the formula to calculate B_wire.
Finally, to find the resulting magnetic field at a point on the axis of the solenoid, we can use the principle of superposition. Since the magnetic fields created by the solenoid and the wire are parallel and pointing in the same direction, we can simply add the magnitudes of the two magnetic fields:
B_resultant = B_solenoid + B_wire.
By substituting the calculated values of B_solenoid and B_wire into this equation, we can find the magnitude of the resulting magnetic field.