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2 years ago

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Two coils of wire are placed close together. Initially, a current of 3.04 A exists in one of the coils, but there is no current

in the other. The current is then switched off in a time of 1.75 x 10-2 s. During this time, the average emf induced in the other coil is 4.29 V. What is the mutual inductance of the two-coil system

1 answer:

Alchen [17]2 years ago

4 0

Answer:

$M=0.0247H$

Explanation:

Given data

$V_{voltage}=4.29V\\I_{current}=3.04A\\t_{time}=1.75*10^{-2}s$

To find

Mutual inductance of the two-coil system

Solution

The mutual inductance given as:

M= (-VΔt)/ΔI

Substitute the given values

So

$M=-\frac{4.29V*1.75*10^{-2}s}{(0-3.04A)}\\ M=0.0247H$

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Answer:

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b) $T=1.9 s$

Explanation:

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The perimeter of a circunference is:

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The horizontal equation of motion will be:

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We can find the angular velocity (ω) from the equation (2):

$\omega=\sqrt{\frac{Tsin(\alpha)}{mr}}=3.31 rad/s$

b) We know that the period is T=2π/ω, therefore:

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