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Chapter 30: Problem 13

. When the current in a long. straight, air-filled solenoid is changing at the rate of \(2000 \mathrm{~A} / \mathrm{s},\) the voltage across the solcnoid is \(0.600 \mathrm{~V}\). The solenoid has 1200 turns and uniform cross-sectional area \(25.0 \mathrm{~mm}^{2}\). Assume that the magnetic field is uniform inside the solenoid and zero outside, so the result \(L=\mu_{0} A N^{2} / l\) (sec Exercise 30.11 ) applics. What is the magnitude \(B\) of the magnctic ficld in the interior of the solenoid when the current in the solenoid is 3,00 A?

Short Answer

Expert verified
The magnitude of the magnetic field in the interior of the solenoid when the current in it is 3 A is \(B = 1.80 \times 10^{-4}\) T.

Step by step solution

01

Understanding Faraday's Law of Electromagnetic Induction

Faraday's Law states that the induced emf is equal to the rate of change of magnetic flux through a coil. It can be written in the formula: \(emf = -L \frac {dI}{dt}\) where \(L\) is inductance and \(\frac {dI}{dt}\) is the rate of change of current.
02

Find the Solenoid's Inductance \(L\)

From Faraday's law, rearrange the equation to solve for \(L\), the inductance: \(L = -\frac {emf}{\frac {dI}{dt}}\). Given an emf of 0.600 V (note that even though the emf is negative in Faraday's law, we are working with magnitudes here) and a rate of change in current of 2000 A/s, the inductance is \(L = -\frac {0.600 V}{2000 A/s} = 0.0003 H\), also known as 300 碌H.
03

Use the Solenoid Self-Inductance Formula

The self-inductance \(L\) of a solenoid can be expressed by the formula \(L=\frac{\mu_{0} A N^{2}}{l}\), where \(\mu_{0}\) is the magnetic permeability of free space, \(A\) is cross-sectional area, \(N\) is the number of turns, and \(l\) is the length of the solenoid. Solving this formula for \(l\) we obtain \(l = \frac{\mu_{0} A N^{2}}{L}\). The cross-sectional area \(A\) is given in \(mm^2\), thus \(A = 25.0 mm^2 = 25.0 \times 10^{-6} m^{2}\). The number of turns \(N\) is given as 1200. The magnetic permeability of free space \(\mu_{0}\) is given as \(4\pi \times 10^{-7} T m/A\), and \(L\) we have calculated in the previous step. Substituting these values into the formula, we compute \(l = \frac {( 4\pi \times 10^{-7} T m/A) (25.0 \times 10^{-6} m^{2})(1200)^2}{0.0003 H}\), yielding \(l = 25.13 m\).
04

Calculating the Magnetic Field \(B\)

The magnetic field \(B\) of a solenoid is expressed by the formula \(B = \mu_{0} \frac {NI}{l}\), where \(N\) is the number of turns, \(I\) is the current, and \(l\) is the length of the solenoid. We rearrange the formula and substitute the given and calculated values, \(B= \frac {\mu_{0} N I}{l}\), where \(\mu_{0} = 4\pi \times 10^{-7} T m/A\), \(N = 1200\), \(I = 3 A\), and \(l = 25.13 m\). From this, we find that \(B = \frac {(4\pi \times 10^{-7} T m/A) \times 1200 \times 3 A}{25.13 m}\) giving \(B = 1.80 \times 10^{-4} T\). This is the magnitude of the magnetic field in the interior of the solenoid when the current in it is 3 A.

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Key Concepts

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

Faraday's Law
Faraday's Law is a foundational principle in the field of electromagnetism, discovered by Michael Faraday. It explains how a change in magnetic flux can induce an electromotive force (emf) in a circuit. The law is mathematically expressed as \( ext{emf} = -L \frac {dI}{dt} \), where \( L \) is the inductance and \( \frac {dI}{dt} \) is the rate of change of current with time.

Faraday's discovery links magnetic fields with electric currents and paved the way for modern electrical engineering. It helps us understand how transformers, inductors, and many types of machines and electric devices operate.
  • The "-" sign in the formula indicates Lenz's Law, which states that the induced emf always opposes the change in current.
  • The greater the change in current or magnetic flux, the greater the induced emf.
This principle is crucial in many applications, especially where electricity generation and transformation is involved, like in power plants and electrical circuits.
Solenoid
A solenoid is essentially a coil of wire designed to create a magnetic field when an electric current passes through it. It is usually cylindrical, and its length is much greater than its diameter, ensuring the magnetic field inside is fairly uniform.

The physical structure and properties of a solenoid make it ideal for controlling magnetic fields and inductance values in electronic circuits. They are widely used in devices like relays, valves, and electromagnets.
  • Inside the solenoid, the magnetic field lines are parallel and closely spaced, indicating a strong and even magnetic field.
  • Outside the solenoid, the field lines spread out and are weak, which is commonly negligible in theoretical calculations.
The strength of the field inside the solenoid depends on the current flowing through it, the number of turns in the coil (\( N \)), and the material around which the coil is wound.
Magnetic Field
The magnetic field in a solenoid is a region where magnetic forces are exerted, and it can be calculated using the formula \( B = \mu_{0} \frac {NI}{l} \). Here, \( B \) is the magnetic field, \( \mu_{0} \) is the permeability of free space, \( N \) is the number of turns, \( I \) is the current, and \( l \) is the length of the solenoid.

In practice, this field is uniform within the solenoid and zero outside due to the contained structure of the coil. This makes solenoids incredibly useful for applications that require predictable magnetic fields.
  • The magnetic permeability \( \mu_{0} = 4\pi \times 10^{-7} \, T \, m/A \) is a constant, easy to remember as it often appears in magnetism equations.
  • The strength of the magnetic field is directly proportional to the current and number of turns and inversely proportional to the solenoid's length.
This understanding is vital for roles like electrical engineering, where magnetic fields need to be manipulated accurately.
Inductance
Inductance is a property of an electrical conductor whereby a change in current through it induces an emf, according to Faraday's Law. In the case of a solenoid, inductance \( L \) can be expressed as \( L = \frac{\mu_{0} A N^{2}}{l} \), where \( A \) is the cross-sectional area, \( N \) is the number of turns, and \( l \) is the length of the solenoid.

Inductance allows the storage of energy in a magnetic field, which is why it is often implemented in circuits to manage change in currents. Inductive components counteract or "smooth" changes in current:
  • A greater number of turns \( N \) or a larger cross-sectional area \( A \) increases inductance, enhancing a solenoid's ability to oppose changes.
  • A longer solenoid means a more spread out field, reducing inductance.
  • The unit of inductance, the henry (H), is crucial in various applications like transformers and inductive sensors.
Understanding inductance is essential for analyzing how circuits respond to signals and changes, making it a key concept in electronics and physics.

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Most popular questions from this chapter

. In Fig. \(30.11 .\) suppose that \(\varepsilon=60.0 \mathrm{~V}, R=240 \Omega .\) and \(L=0.160 \mathrm{H}\). With switch \(S_{2}\) open, switch \(S_{1}\) is left closed until a constant currcnt is cstablishcd. Then \(S_{2}\) is closcd and \(S_{1}\) opcned, taking the battery out of the circuit. (a) What is the initial current in the resistor, just after \(S_{2}\) is closed and \(S_{1}\) is opened? (b) What is the current in the resistor at \(t=4.00 \times 10^{-4} \mathrm{~s} ?\) (c) What is the potcntial differcnoe betwecn points \(b\) and \(c\) at \(t=4.00 \times 10^{-4} \mathrm{~s}\) ? Which point is at a higher potcntial? (d) How long does it take the current to decrease to half its initial value?

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns (see Example 30.1 ). The inner solenoid is \(25.0 \mathrm{~cm}\) long and has a diameter of \(2.00 \mathrm{~cm}\). At a certain time, the current in the inner solenoid is \(0.120 \mathrm{~A}\) and is increasing at a rate of \(1.75 \times 10^{3} \mathrm{~A} / \mathrm{s}\). For this time, calculate: (a) the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

Two coils have mutual inductance \(M=3.25 \times 10^{4}\) H. The current \(i_{1}\) in the first coil increases at a uniform rate of \(830 \mathrm{~A} / \mathrm{s}\). (a) What is the magnitude of the induced emf in the second coil? Is it constant? (b) Suppose that the current described is in the second coil rather than the first. What is the magnitude of the induced emf in the first coil?

It is possible to make your own inductor by winding wire around a cylinder, such as a pcncil. Assume you have a spool of AWG 20 copper wire, which has a diameter of \(0.812 \mathrm{~mm}\). (a) Estimate the diameter of a pencil. (b) Estimate how many times can you tightly wrap AWG 20 copper wire around a pencil to form a solenoid with a length of \(4.0 \mathrm{~cm}\). (c) Estimate the inductance of this solcnoid by assuming the magnetic field inside is constant. (d) If a current of 1.0 A flows through this solenoid, how much magnetic energy will be stored inside?

DATA To investigate the properties of a large industrial solenoid, you connect the solenoid and a resistor in series with a battery. Switches allow the battery to be replaced by a short circuit across the solenoid and resistor. Therefore Fig. 30.11 applies, with \(R=R_{\text {ext }}+R_{L}\). where \(R_{L}\) is the resistance of the solenoid and \(R_{\text {ext }}\) is the resistance of the series resistor. With switch \(S_{2}\) opcn, you close switch \(S_{1}\) and kecp it closed until the current \(i\) in the solenoid is constant (Fig. 30.11 ). Then you close \(S_{2}\) and open \(S_{1}\) simultaneously, using a rapid- response switching mechanism. With high-speed clectronics you measure the time \(t_{\text {balf }}\) that it takes for the current to decrease to half of its initial value. You repeat this measurement for several values of \(R_{\text {ext }}\) and obtain these results: \begin{tabular}{l|lllllll} \(R_{\text {ext }}(\Omega)\) & 3.0 & 4.0 & 5.0 & 6.0 & 7.0 & 8.0 & 10.0 & 12.0 \\\ \hline\(t_{\text {half }}(\mathrm{s})\) & 0.735 & 0.654 & 0.589 & 0.536 & 0.491 & 0.453 & 0.393 & 0.347 \end{tabular} (a) Graph your data in the form of \(1 / t_{\text {halt }}\) versus \(R_{e x l}\). Explain why the data points plotted this way fall close to a straight line. (b) Use your graph from part (a) to calculate the resistance \(R_{L}\) and inductance \(L\) of the solenoid. (c) If the current in the solenoid is \(20.0 \mathrm{~A}\), how much energy is stored there? At what rate is electrical energy being dissipated in the resistance of the solenoid?
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