A long solenoid ( radius = 1.8 cm) has a current of a 0.33 A in its winding. A long wire carrying a current of 22 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?
To find the magnitude of the resulting magnetic field at a point on the axis of the solenoid, we can use the principle of superposition, which states that the total magnetic field at a given point is the vector sum of the magnetic fields produced by each individual source.
In this case, we have two sources of magnetic field: the solenoid and the long wire.
The magnetic field produced by a long solenoid can be calculated using the formula:
B_solenoid = (μ_0 * n * I) / L,
where B_solenoid is the magnetic field, μ_0 is the permeability of free space (4π x 10^-7 T·m/A), n is the number of turns per unit length of the solenoid, I is the current flowing through the solenoid, and L is the length of the solenoid (which we'll assume to be very long).
Given that the radius of the solenoid is 1.8 cm, we can calculate the number of turns per unit length 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.
Since the solenoid is assumed to be very long, we can consider it as having a finite length compared to its radius. In this case, we can approximate the length as the length of the solenoid itself (ignoring the end effects).
Now, let's calculate the values one by one:
The radius of the solenoid is given as 1.8 cm, which is equivalent to 0.018 m.
The current flowing through the solenoid is given as 0.33 A.
The distance between the wire and the axis of the solenoid is given as 1.2 cm, which is equivalent to 0.012 m.
The current flowing through the wire is given as 22 A.
Now, we can calculate the number of turns per unit length using the given radius of the solenoid:
n = N / L,
Since the length of the solenoid is not given, we can assume it to be 1 meter for simplicity.
n = N / L = (2π * 0.018 m) / 1 m ≈ 0.113 m^-1.
Next, we calculate the magnetic field produced by the solenoid using the formula:
B_solenoid = (μ_0 * n * I) / L,
B_solenoid = (4π × 10^-7 T·m/A * 0.113 m^-1 * 0.33 A) / 1 m,
B_solenoid ≈ 4.14 × 10^-6 T.
Now, let's calculate the magnetic field produced by the wire. The formula to calculate the magnetic field produced by a long straight wire is:
B_wire = (μ_0 * I) / (2π * r),
where B_wire is the magnetic field, μ_0 is the permeability of free space, I is the current flowing through the wire, and r is the distance from the wire.
Using the given values:
B_wire = (4π × 10^-7 T·m/A * 22 A) / (2π * 0.012 m),
B_wire ≈ 3.67 × 10^-4 T.
Now, to find the resultant magnetic field at a point on the axis of the solenoid, we add the magnetic fields produced by the solenoid and the wire:
B_resultant = B_solenoid + B_wire,
B_resultant ≈ 4.14 × 10^-6 T + 3.67 × 10^-4 T,
B_resultant ≈ 3.71 × 10^-4 T.
Therefore, the magnitude of the resulting magnetic field at a point on the axis of the solenoid is approximately 3.71 × 10^-4 T.