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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inductance
Inductance is a fundamental property of electrical circuits that relates to the ability of a coil to store energy in a magnetic field. It is often denoted by the symbol \( L \) and is measured in henrys (H), named after Joseph Henry, an American scientist.
This property is especially crucial in circuits that experience varying currents because the inductance creates a sort of "resistance" to changes in current. This is due to the energy stored in the magnetic field surrounding the coil.
  • A higher inductance means more energy can be stored in the magnetic field, leading to a greater opposition to changes in current.
  • Inductive components become particularly important in AC (alternating current) circuits.
Understanding inductance helps us comprehend how electric circuits mitigate fluctuations in current, maintaining a stable performance under dynamic conditions.
Ohm's Law
Ohm's Law is a cornerstone principle in the study of electric circuits. It provides a simple relationship between voltage, current, and resistance in a circuit.
This law is usually expressed by the formula: \( V = IR \), where \( V \) is the voltage (in volts), \( I \) is the current (in amperes), and \( R \) is the resistance (in ohms).
  • In practical terms, Ohm's Law allows you to calculate any one of these variables if you know the other two.
  • For instance, the maximum current through a circuit can be found when the voltage and resistance are given, using the reformatted equation \( I = \frac{V}{R} \).
This law is crucial for designing and analyzing circuits, ensuring that components are properly matched for efficient operation.
Energy Stored in Magnetic Field
Electric circuits that include inductive components have the capacity to store energy in magnetic fields. This energy storage is vital because it influences how the circuit behaves, especially under varying conditions.
The energy stored in a magnetic field can be calculated using the formula: \( W = \frac{1}{2} L I^2 \), where \( W \) is the energy (in joules), \( L \) is the inductance (in henrys), and \( I \) is the current (in amperes).
  • This formula shows that the energy stored is proportional to both the inductance and the square of the current flowing through the coil.
  • This stored energy can be used to smooth out current changes, preventing abrupt spikes that could damage components.
Understanding this concept helps ensure that circuits are both stable and safe during operation.
Time Constant of RL Circuit
In an RL circuit, which includes both a resistor (R) and an inductor (L), the time constant is a measure of how quickly the circuit responds to changes in voltage. This time constant, often denoted by \( \tau \), provides insight into the dynamic behavior of the circuit.
The time constant is calculated using the formula: \( \tau = \frac{L}{R} \), where \( L \) is the inductance and \( R \) is the resistance.
  • The time constant gives an indication of how long it takes for the current to reach approximately 63% of its final value in response to a step change in voltage.
  • A larger time constant means the circuit responds more slowly, which can be beneficial in applications where gradual changes are preferred.
Knowing the time constant helps in designing circuits that require specific response times, ensuring they operate reliably and efficiently.

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Most popular questions from this chapter

A metal loop has an area of \(0.100 \mathrm{~m}^{2}\) and is placed flat on the ground. There is a uniform magnetic field pointing due west, as shown in the figure. This magnetic field initially has a magnitude of \(0.123 \mathrm{~T}\), which decreases steadily to \(0.075 \mathrm{~T}\) during a period of \(0.579 \mathrm{~s}\). Find the potential difference induced in the loop during this time.

Two parallel conducting rails with negligible resistance are connected at one end by a resistor of resistance \(R\), as shown in the figure. The rails are placed in a magnetic field \(\vec{B}_{\text {ext }},\) which is perpendicular to the plane of the rails. This magnetic field is uniform and time independent. The distance between the rails is \(\ell\). A conducting rod slides without friction on top of the two rails at constant velocity \(\vec{v}\). a) Using Faraday's Law of Induction, calculate the magnitude of the potential difference induced in the moving rod. b) Calculate the magnitude of the induced current in the \(\operatorname{rod}, i_{\text {ind }}\). c) Show that for the rod to move at a constant velocity as shown, it must be pulled with an external force, \(\vec{F}_{\mathrm{ext}},\) and calculate the magnitude of this force. d) Calculate the work done, \(W_{\text {ext }},\) and the power generated, \(P_{\text {ext }}\), by the external force in moving the rod. e) Calculate the power used (dissipated) by the resistor, \(P_{\mathrm{R}}\). Explain the correlation between this result and those of part (d).

Having just learned that there is energy associated with magnetic fields, an inventor sets out to tap the energy associated with the Earth's magnetic field. What volume of space near Earth's surface contains \(1 \mathrm{~J}\) of energy, assuming the strength of the magnetic field to be \(5.0 \cdot 10^{-5} \mathrm{~T} ?\)

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.

A long solenoid with length \(3.0 \mathrm{~m}\) and \(n=290\) turns \(/ \mathrm{m}\) carries a current of \(3.0 \mathrm{~A} .\) It stores \(2.8 \mathrm{~J}\) of energy. What is the cross-sectional area of the solenoid?

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