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A magnetic field (B=3.2x10^-3 T) passes perpendicular through a circular loop of wire ( radius= 5.0 cm) what is the magnetic flux through the hoop?

2 Answers

1 vote

Final answer:

The magnetic flux through a circular loop of wire is found by multiplying the magnetic field strength by the area of the loop, considering the magnetic field is perpendicular to the loop's plane.

Step-by-step explanation:

The subject of this question is Physics, specifically related to the topic of magnetic fields and magnetic flux. To find the magnetic flux through a circular loop of wire given the magnetic field strength and the radius of the loop, we can use the formula Φ = B × A × cos(θ). Since the magnetic field is perpendicular to the loop, θ is 0 degrees, and cos(0) = 1. Therefore, the magnetic flux Φ is simply the product of the magnetic field B and the area A of the loop.

To calculate the area A of a circle, we use the formula A = πr^2, where r is the radius of the circle. For a radius of 5.0 cm, the area A is π(0.05 m)^2. Multiplying the area by the given magnetic field B = 3.2 x 10^-3 T, we get the magnetic flux.

User Jmucchiello
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4 votes

Answer:

25.12 x 10^-6 T/m^2

Step-by-step explanation:

Magnetic flux is

fi = B * A * cos theta

as filed passes perepndicular theta is 90degrees and cos theta is 1

B=3.2x10^-3 T

A = pi * r ^2

A = 3.14 * (0,05m )^2

A = 0,00785

A = 7,85 m 10^-3 m2

fi = 3.2x10^-3 T * 7,85 m 10^-3 m2

fi = 25.12 x 10^-6 T/m^2

User Sccs
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4.9k points