FIGURE 23-5
-A rectangular loop of wire carrying a current of 2.0 A that can rotate is placed in a magnetic field with a strength of 0.40 T as shown in Figure 23-5. The length of the loop L is 0.16 m and its width w is 0.040 m. What is the magnetic flux through the loop when the plane of the loop is perpendicular to the magnetic field?

Question

FIGURE 23-5 
-A rectangular loop of wire carrying a current of 2.0 A that can rotate is placed in a magnetic field with a strength of 0.40 T as shown in Figure 23-5. The length of the loop L is 0.16 m and its width w is 0.040 m. What is the magnetic flux through the loop when the plane of the loop is perpendicular to the magnetic field?

Quizplus · Accepted Answer

The magnetic flux (Φ) through the loop is given by Φ = B * A * cos(θ), where B is the magnetic field strength, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the plane of the loop. When the plane of the loop is perpendicular to the magnetic field, θ = 0 degrees, and cos(θ) = 1. The area A = L * w = 0.16 m * 0.040 m = 0.0064 m². Therefore, Φ = 0.40 T * 0.0064 m² = 2.56 × 10^-3 T∙m², which rounds to 2.6 × 10^-3 T∙m².

FIGURE 23-5 -A Rectangular Loop of Wire Carrying a Current of 2.0