A single-turn square loop of side L is centered on the axis of a long solenoid. In addition, the plane of the square loop is perpendicular to the axis of the solenoid. The solenoid has 1490 turns per meter and a diameter of 6.00 cm, and carries a current of 2.46 A. Find the magnetic flux through the loop in each of the following.
a) L = 3cm
b) L = 6cm
c) L = 12cm
I found a to be 4.15e-6, b to be 1.66e-5 and c to be 6.63e-5, but only a is right. the answers are within ten percent of 1.2e-5 and are the same answer but I can't find what they should be. Please someone help soon.
To find the magnetic flux through the loop, we can use the formula:
Φ = B * A * cosθ
Where Φ is the magnetic flux, B is the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop.
In this case, the solenoid produces a uniform magnetic field along its axis. The magnetic field inside the solenoid can be calculated using the formula:
B = μ₀ * n * I
Where μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current.
Given that the solenoid has 1490 turns per meter and carries a current of 2.46 A, we can calculate the magnetic field B using the above formula.
Next, we need to calculate the area A of the square loop. The area of a square can be calculated using the formula:
A = L²
Where L is the length of the side of the square loop.
Finally, since the square loop is perpendicular to the axis of the solenoid, the angle θ between the magnetic field and the normal to the loop is 90 degrees. Therefore, cos90° = 0.
Plugging in the values into the formula, we get:
Φ = B * A * cosθ = B * L² * 0 = 0
This means that the magnetic flux through the loop is zero for all values of L.
So, the correct answer for all three values of L is 0, not the values you calculated.
To find the magnetic flux through the loop, we can use the formula:
Φ = B * A
Where:
Φ is the magnetic flux,
B is the magnetic field,
A is the area of the loop.
First, let's calculate the magnetic field at the center of the solenoid using the formula:
B = μ₀ * n * I
Where:
B is the magnetic field,
μ₀ is the magnetic constant (4π × 10^-7 T·m/A),
n is the number of turns per unit length (1490 turns/m),
I is the current in the solenoid (2.46 A).
Next, we need to determine the area of the loop. Since the square loop is centered on the axis, we can consider the side length as the diameter of a circle. So, the area of the loop is:
A = (π/4) * L²
Finally, we can calculate the magnetic flux for each of the given side lengths:
a) L = 3 cm:
A = (π/4) * (0.03 m)²
B = (4π × 10^-7 T·m/A) * (1490 turns/m) * (2.46 A)
Φ = B * A
b) L = 6 cm:
A = (π/4) * (0.06 m)²
B = (4π × 10^-7 T·m/A) * (1490 turns/m) * (2.46 A)
Φ = B * A
c) L = 12 cm:
A = (π/4) * (0.12 m)²
B = (4π × 10^-7 T·m/A) * (1490 turns/m) * (2.46 A)
Φ = B * A
Using these calculations, we can determine the magnetic flux for each case.
Hmmm. A is right, you say.
The area of L^2 in a is 9cm^2
the area of the solenoid is 9PI
Area outside the solenoid does not contain the inner field.
b. Flux=4.15E-6* 9PI/9=4.15E-6*PI=1.304E-5
That is 8 percent higher than 1.2E-5
Now, larger L does not contain any more flux, as the flux is contained within the solnoid
Think about this, I may not be understanding the problem.