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Chapter 29: Problem 52

An emf of \(20.0 \mathrm{~V}\) is applied to a coil with an inductance of \(40.0 \mathrm{mH}\) and a resistance of \(0.500 \Omega\). a) Determine the energy stored in the magnetic field when the current reaches one fourth of its maximum value. b) How long does it take for the current to reach this value?

Short Answer

Expert verified
Answer: The energy stored in the magnetic field when the current reaches one-fourth of its maximum value is 2.00 J, and it takes 27.73 ms for the current to reach that value.

Step by step solution

01

Determine the maximum current

To find the maximum current in the coil, use Ohm's law with the provided emf and resistance: $$I_{max} = \frac{E}{R}$$ Plugging in the given values, we have: $$I_{max} = \frac{20.0\mathrm{~V}}{0.500\Omega} = 40.0\mathrm{~A}$$
02

Find the current as a function of time

Considering the inductance and resistance in the circuit, the current as a function of time is given by: $$I(t) = I_{max}(1 - e^{-\frac{Rt}{L}})$$ Now, we want to find when the current reaches one-fourth of its maximum value: $$I_{\frac{1}{4}} = \frac{1}{4} I_{max}$$
03

Determine the time it takes to reach one-fourth of the maximum current

Equating the above expressions, we have: $$\frac{1}{4} I_{max} = I_{max}(1 - e^{-\frac{Rt}{L}})$$ Dividing both sides by \(I_{max}\) and rearranging, we obtain: $$e^{-\frac{Rt}{L}} = 1 - \frac{1}{4} \Rightarrow -\frac{Rt}{L} = \ln\left(\frac{3}{4}\right)$$ Solving for \(t\), we have: $$t = -\frac{L}{R}\ln\left(\frac{3}{4}\right)$$ Plugging in the given values for \(R\) and \(L\), we find: $$t = -\frac{40.0\mathrm{~mH}}{0.500\Omega}\ln\left(\frac{3}{4}\right) = 27.73\mathrm{~ms}$$
04

Calculate the energy stored in the magnetic field

The energy stored in the magnetic field when the current reaches one-fourth of its maximum value, is given by: $$W = \frac{1}{2}LI_{\frac{1}{4}}^2$$ Using the calculated values from steps 1 and 3, we have: $$W = \frac{1}{2}(40.0\mathrm{~mH})(10.0\mathrm{~A})^2 = 2.00\mathrm{~J}$$ So the energy stored in the magnetic field when the current reaches one-fourth of its maximum value is \(2.00\mathrm{~J}\), and it takes \(27.73\mathrm{~ms}\) for the current to reach that value.

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Most popular questions from this chapter

Large electric fields are certainly a hazard to the human body, as they can produce dangerous currents, but what about large magnetic fields? A man \(1.80 \mathrm{~m}\) tall walks at \(2.00 \mathrm{~m} / \mathrm{s}\) perpendicular to a horizontal magnetic field of \(5.0 \mathrm{~T} ;\) that is, he walks between the pole faces of a very big magnet. (Such a magnet can, for example, be found in the National Superconducting Cyclotron Laboratory at Michigan State University.) Given that his body is full of conducting fluids, estimate the potential difference induced between his head and feet.

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At Los Alamos National Laboratories, one means of producing very large magnetic fields is known as the EPFCG (explosively-pumped flux compression generator), which is used to study the effects of a high-power electromagnetic pulse (EMP) in electronic warfare. Explosives are packed and detonated in the space between a solenoid and a small copper cylinder coaxial with and inside the solenoid, as shown in the figure. The explosion occurs in a very short time and collapses the cylinder rapidly. This rapid change creates inductive currents that keep the magnetic flux constant while the cylinder's radius shrinks by a factor of \(r_{\mathrm{i}} / r_{\mathrm{f}}\). Estimate the magnetic field produced, assuming that the radius is compressed by a factor of 14 and the initial magnitude of the magnetic field, \(B_{i}\), is \(1.0 \mathrm{~T}\).

A short coil with radius \(R=10.0 \mathrm{~cm}\) contains \(N=30.0\) turns and surrounds a long solenoid with radius \(r=8.00 \mathrm{~cm}\) containing \(n=60\) turns per centimeter. The current in the short coil is increased at a constant rate from zero to \(i=2.00 \mathrm{~A}\) in a time of \(t=12.0 \mathrm{~s}\). Calculate the induced potential difference in the long solenoid while the current is increasing in the short coil.
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