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Chapter 27: Problem 61

When a nonideal \(1.0-\mathrm{H}\) inductor is short-circuited, its magnetic energy drops to one-fourth of its original value in \(3.6 \mathrm{s}\). What is its resistance?

Short Answer

Expert verified
The resistance of the inductor is approximately 0.42Ω.

Step by step solution

01

Expression for energy decay

Firstly, using the equation of energy decay over time, we have \(U(t) = U(0) e^{-t/τ}\). We know that the magnetic energy drops to one-fourth of its original value in 3.6s. So, we can equate \(\frac{1}{4}U(0)\) to \(U(0)e^{-t/τ}\) and solve for \(τ\). This simplifies to \(e^{-3.6/τ} = \frac{1}{4}\)
02

Calculate value of time constant

Taking natural logarithm on both sides gives \(-\frac{3.6}{τ} = ln(1/4)\). Solving for τ yields τ = -3.6 / ln(1/4) = 2.4s.
03

Calculate resistance

Now, knowing the value of τ, we can use the relation τ = L/R to solve for R. If we take L as 1.0H (Given), Then, R = L/τ = 1.0/2.4 = 0.42Ω.

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Most popular questions from this chapter

A flip coil is used to measure magnetic fields. It's a small coil placed with its plane perpendicular to a magnetic field, and then flipped through \(180^{\circ} .\) The coil is connected to an instrument that measures the total charge \(Q\) that flows during this process. If the coil has \(N\) turns, area \(A,\) and resistance \(R,\) show that the field strength is \(B=Q R / 2 N A.\)

A battery, switch, resistor, and inductor are connected in series. When the switch is closed, the current rises to half its steady state value in 1.0 ms. How long does it take for the magnetic energy in the inductor to rise to half its steady-state value?

Find the magnetic flux through a 5.0 -cm-diameter circular loop oriented with the loop normal at \(30^{\circ}\) to a uniform 80 -mT magnetic field.

Chapter 26 claimed that a static magnetic field cannot change the energy of a charged particle. Is this true of a changing magnetic field? Discuss.

Find the magnetic-field strength in a region where the magneticenergy density is \(7.8 \mathrm{J} / \mathrm{cm}^{3} .\)
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