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The 2.7-cm-diameter solenoid in (Figure 1) passes through the center of a 6.0-cm-diameter loop. The magnetic field inside the solenoid is 0.30 T .

A. What is the magnetic flux through the loop when it is perpendicular to the solenoid?
B. What is the magnetic flux through the loop when it is tilted at a 60 angle?

User Aabujamra
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Final answer:

The magnetic flux through the loop when it is perpendicular to the solenoid is 8.478 cm^2, and when it is tilted at a 60° angle, the magnetic flux is 7.352 cm^2.

Step-by-step explanation:

When the loop is perpendicular to the solenoid, the magnetic flux through the loop can be calculated using the formula: φ = B * A, where B is the magnetic field and A is the area of the loop. In this case, the diameter of the loop is 6.0 cm, so the radius is 3.0 cm. The area of the loop is then calculated as A = π * r2 = 3.14 * (3.0 cm)2 = 28.26 cm2. Substituting the values into the formula, we get: φ = 0.30 T * 28.26 cm2 = 8.478 cm2



When the loop is tilted at a 60° angle, the magnetic flux through the loop can still be calculated using the same formula. However, since the loop is tilted, we need to consider the component of the magnetic field that is perpendicular to the loop. Assuming the magnetic field is still 0.30 T, the perpendicular component of the magnetic field can be calculated as B_perpendicular = B * sin(60°) = 0.30 T * sin(60°) = 0.26 T. Substituting this value into the formula, we get: φ = 0.26 T * A = 0.26 T * 28.26 cm2 = 7.352 cm2

User CReaTuS
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