Question

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 1350 turns per meter and a diameter of 5.60 cm, and carries a current of 2.80 A.Part A: Find the magnetic flux through the loop when L= 2.75 cm .Part B: Find the magnetic flux through the loop when L= 5.60 cm .Part C: Find the magnetic flux through the loop when L= 13.5 cm .

Answer 1

Answer:

(A) $\Phi_{\rm B} = 4.9\times 10^{-4}$

(B) $\Phi_{\rm B} = 1.1\times 10^{-5}$

(C) $\Phi_{\rm B} = 1.1\times 10^{-5}$

Explanation:

The magnetic flux is equal to the area multiplied by the magnetic field passes through that area.

$\Phi_{\rm B} = \vec{B}\vec{A} = BA\cos{\theta}$

Since the loop is perpendicular to the axis of the solenoid, cosine term is equal to 1.

The magnetic field created by a solenoid can be found by Ampere's Law:

$\int \vec{B}d\vec{l} = \mu_0 I_{enc}$

The integral in the left-hand side of the equation is unnecessary if we choose the imaginary Amperian loop a rectangle whose one side is inside the solenoid and the other side is outside. Let us denote the length of the loop which is inside the solenoid as 'd'.

$Bd = \mu_0 I_{enc}\\B = \frac{\mu_0 (2.8\times 1350d)}{d} = \mu_0 (3.78\times 10^3)$

Here, the length of the loop 'd' is multiplied by 1350 (turns per meter) to find the number of turns per 'd'. Each turn carries a current of 2.8 A.

Part A:

The area of the loop with side length L = 2.75 cm is

$A = 4\times 2.75 \times 10^{-2} = 0.11 ~{\rm m^2}$

$\Phi_{\rm B} = BA = 415.8\mu_0 = 4.89\times 10^{-4}$

Part B:

When L = 5.6 cm, the area of the loop is greater than the area of the solenoid. Since the magnetic field is only present inside the solenoid, the magnetic field will only go through the area of the solenoid.

$\Phi_{\rm B} = BA = B\pi r^2 = \mu_0(3780)\pi (2.8\times 10^{-2})^2 = 9.31\mu_0 = 1.11\times 10^{-5}$

Part C:

When L is greater than the diameter of the solenoid, the magnetic flux is constant and equal to the magnetic field times the area of the solenoid. So, the flux is equal to the flux in Part B.