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 determine the magnitude of the resulting magnetic field at a point on the axis of the solenoid, we can use Ampere's Law. Ampere's Law states that the line integral of the magnetic field around a closed loop is equal to the product of the current passing through the loop and the permeability of free space.
1. First, we need to calculate the magnetic field at a distance of 1.2 cm from the axis of the solenoid due to the wire carrying a current of 22 A.
To determine this, we can use the formula for the magnetic field due to a current-carrying wire. The formula is given by:
B = (μ₀ * I) / (2π * r)
Where:
B is the magnetic field.
μ₀ is the permeability of free space (4π x 10^-7 T.m/A).
I is the current in the wire.
r is the distance from the wire.
Substituting the given values:
I = 22 A
r = 1.2 cm = 0.012 m
μ₀ = 4π x 10^-7 T.m/A
B = (4π x 10^-7 * 22) / (2π * 0.012)
B = 0.36667 T
2. Next, we calculate the magnetic field at a point on the axis of the solenoid due to the solenoid itself.
The magnetic field inside a solenoid is given by:
B = (μ₀ * n * I) / l
Where:
B is the magnetic field.
μ₀ is the permeability of free space (4π x 10^-7 T.m/A).
n is the number of turns per unit length.
I is the current in the solenoid winding.
l is the length of the solenoid.
Given values:
I = 0.33 A
r = 1.8 cm = 0.018 m
n = 1 (assuming there is only one layer of winding)
l is not provided.
Here, we need to consider that the length of the solenoid should be large compared to its radius. Consequently, the magnetic field inside the solenoid near its axis is approximately equal to the magnetic field at the center of a circular loop of the same radius (inside the solenoid).
So, we can use the magnetic field formula for a circular loop at its center:
B = (μ₀ * I) / (2 * r)
Substituting the given values:
I = 0.33 A
r = 0.018 m
μ₀ = 4π x 10^-7 T.m/A
B = (4π x 10^-7 * 0.33) / (2 * 0.018)
B ≈ 0.009 T
3. Finally, to find the magnitude of the resulting magnetic field at a point on the axis of the solenoid, we add up the magnetic fields due to the wire and the solenoid together since they are parallel and in the same direction:
B_total = B_wire + B_solenoid
B_total = 0.36667 T + 0.009 T
B_total ≈ 0.37567 T
Therefore, the magnitude of the resulting magnetic field at a point on the axis of the solenoid is approximately 0.37567 T.
To find the magnitude of the resulting magnetic field at a point on the axis of the solenoid, we can use the Biot-Savart law.
The Biot-Savart law states that the magnetic field at a point due to a current-carrying wire is proportional to the current and inversely proportional to the distance from the wire.
The formula for the magnetic field due to a current-carrying wire is:
B = (μ₀ * I) / (2π * r)
Where:
B is the magnetic field
μ₀ is the permeability of free space (4π * 10^-7 T*m/A)
I is the current
r is the distance from the wire
We will calculate the magnetic field due to the solenoid and the wire separately, and then add them together.
First, let's calculate the magnetic field due to the solenoid:
Given:
Radius of solenoid (r₁) = 1.8 cm = 0.018 m
Current in solenoid (I₁) = 0.33 A
Using the Biot-Savart law, we can calculate the magnetic field due to the solenoid:
B₁ = (μ₀ * I₁) / (2π * r₁)
Next, let's calculate the magnetic field due to the wire:
Given:
Distance from the wire (r₂) = 1.2 cm = 0.012 m
Current in wire (I₂) = 22 A
Using the Biot-Savart law, we can calculate the magnetic field due to the wire:
B₂ = (μ₀ * I₂) / (2π * r₂)
Finally, we can add the two magnetic fields together to find the resulting magnetic field at a point on the axis of the solenoid:
B = B₁ + B₂
Let's substitute the values into the formulas and calculate the magnetic fields.