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. A single strand of wire of adjustable length is wound around the circumference of a round balloon that has a diameter of 21.0 cm. A uniform magnetic field with a magnitude of 1.40 T is perpendicular to the plane of the loop. If the balloon is blown up such that its diameter and the diameter of the wire loop increase to 31.0 cm in 0.048s, what is the magnitude of the average value of the emf induced in the loop

Respuesta :

Answer:

EMF = 1.1929 V

Explanation:

In this case we need to use the following expression:

EMF = B *ΔA / t   (1)

Where:

B: Magnetic field

ΔA: Difference of area of the balloon

t: time

To get the area, we have the innitial diameter of the balloon, and the expanded diameter when the balloon is blown up, so, assuming the circunference of the balloon, the area can be determined with:

A = π r²   (2)

So, let's calculate the Area of the balloon before and after the blown. Remember to use the units in meter (I just divided the diameter by 2 to get the ratio, and then divide by 100 to convert the units of cm to m)

A₁ = π * (0.105)² = 0.0346 m²

A₂ = π * (0.155)² = 0.0755 m²

Now that we have both areas, let's calculate ΔA:

ΔA = A₂ - A₁

ΔA = 0.0755 - 0.0346 = 0.0409 m²

Finally, we can determine the value of the EMF, using (1):

EMF = 1.4 * 0.0409 / 0.048

EMF = 1.1929 V

Hope this helps

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