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Chapter 32: Problem 1

A solenoid is \(40 \mathrm{~cm}\) long, has a cross-sectional area of \(8.0 \mathrm{~cm}^{2}\), and is wound with 300 turns of wire that carry a current of \(1.2 \mathrm{~A}\). The relative permeability of its iron core is 600 . Compute (a) \(B\) for an interior point and \((b)\) the flux through the solenoid. (a) From, in air $$ \begin{array}{l} B_{0}=\frac{\mu_{0} N I}{L}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(300)(1.2 \mathrm{~A})}{0.40 \mathrm{~m}}=1.13 \mathrm{mT} \\ \text { and so } \quad B=k_{M} B_{0}=(600)\left(1.13 \times 10^{-3} \mathrm{~T}\right)=0.68 \mathrm{~T} \end{array} $$ (b) Because the field lines are perpendicular to the cross section of the solenoid, $$ \Phi_{M}=B_{\perp} A=B A=(0.68 \mathrm{~T})\left(8.0 \times 10^{-4} \mathrm{~m}^{2}\right)=54 \mu \mathrm{Wb} $$

Short Answer

Expert verified
(a) \(B = 0.68 \text{ T}\); (b) Flux \(\Phi_M = 54 \mu \text{Wb}\).

Step by step solution

01

Calculate Magnetic Field in Air

To find the magnetic field in air at the interior point of the solenoid, use the formula \( B_0 = \frac{\mu_0 N I}{L} \). Here, \( \mu_0 = 4\pi \times 10^{-7} \) T路m/A is the permeability of free space, \( N = 300 \) is the number of turns, \( I = 1.2 \) A is the current, and \( L = 0.40 \) m is the length of the solenoid. Plugging in these values:\[ B_0 = \frac{(4\pi \times 10^{-7})(300)(1.2)}{0.40} = 1.13 \times 10^{-3} \text{ T (or 1.13 mT)} \].
02

Calculate Magnetic Field in Solenoid with Iron Core

The actual magnetic field in the solenoid is strengthened by the iron core, and is calculated by multiplying the field in air by the relative permeability, \( k_M = 600 \). Use the formula \( B = k_M B_0 \), where \( B_0 = 1.13 \times 10^{-3} \) T:\[ B = 600 \times 1.13 \times 10^{-3} = 0.68 \text{ T} \].
03

Calculate Magnetic Flux Through the Solenoid

To find the magnetic flux \( \Phi_M \) through the solenoid, use the formula \( \Phi_M = B A \) where \( B = 0.68 \) T is the magnetic field and \( A = 8.0 \times 10^{-4} \) m虏 is the cross-sectional area. The field lines are perpendicular to the cross-section, simplifying the calculation:\[ \Phi_M = 0.68 \times 8.0 \times 10^{-4} = 54 \times 10^{-6} \text{ Wb} \text{ (or } 54 \mu\text{Wb)} \].

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Key Concepts

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

Magnetic Field
A magnetic field is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. In the context of a solenoid, the magnetic field inside is created by electric current flowing through its coils. The formula to calculate the magnetic field in air for a solenoid is given by:
\[ B_0 = \frac{\mu_0 N I}{L} \]
Where:
  • \( \mu_0 \) is the permeability of free space (\(4\pi \times 10^{-7} \) T路m/A)
  • \( N \) is the number of turns in the solenoid
  • \( I \) is the current flowing through the solenoid
  • \( L \) is the length of the solenoid
The magnetic field in air for the solenoid, as calculated, is \(1.13 \text{ mT} \). If an iron core fills the solenoid, the magnetic field becomes enhanced.
Magnetic Flux
The concept of magnetic flux helps us understand how much magnetic field passes through a defined area. It's a measure of the number of magnetic field lines passing through a surface. The formula to calculate the magnetic flux \( \Phi_M \) through the solenoid is:
\[ \Phi_M = B A \]
Where:
  • \( B \) is the magnetic field (\(0.68 \text{ T} \))
  • \( A \) is the cross-sectional area of the solenoid (\(8.0 \times 10^{-4} \text{ m}^2\))
The calculated magnetic flux is about \(54 \mu \text{Wb} \), and this is because the field lines are perpendicular to the solenoid's cross-section. This simplification allows direct multiplication of \( B \) and \( A \).
Relative Permeability
Relative permeability is the ratio of the permeability of a specific medium (like the iron core of a solenoid) to the permeability of free space (\( \mu_0 \)). It indicates how much better the medium can support the formation of a magnetic field within it compared to air. In simple terms, the higher the relative permeability, the more the material allows magnetic field lines to pass through easily.
When considering the solenoid with an iron core, the formula to find the resultant magnetic field is:
\[ B = k_M B_0 \]
Here, \( k_M \) is the relative permeability, and \( B_0 \) is the magnetic field in air. With a relative permeability of 600 for iron in this example, the magnetic field inside the solenoid becomes significantly enhanced to \(0.68 \text{ T} \) from the air value. Therefore, the iron core boosts the magnetic field strength by 600 times its air value, demonstrating the solenoid's enhanced performance.

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Most popular questions from this chapter

A flat coil with a radius of \(8.0 \mathrm{~mm}\) has 50 turns of wire. It is placed in a magnetic field \(B=0.30 \mathrm{~T}\) in such a way that the maximum flux goes through it. Later, it is rotated in \(0.020 \mathrm{~s}\) to a position such that no flux goes through it. Find the average emf induced between the terminals of the coil.

[III] The metal bar of length \(L\), mass \(m\), and resistance \(R\) depicted in slides without friction on a rectangular circuit composed of resistanceless wire resting on an inclined plane. There is a vertical uniform magnetic field \(\overrightarrow{\mathbf{B}}\). Find the terminal speed of the bar (that is, the constant speed it attains). Gravity pulls the bar down the incline as shown in \(\underline{\text { Fig. }} 32-7(b)\). Induced current flowing in the bar interacts with the field so as to retard this motion. Because of the motion of the bar in the magnetic field, an emf is induced in the bar: $$ \mathscr{E}=(B l v)_{\perp}=B L(v \cos \theta) $$ This causes a current $$ I=\frac{\mathrm{emf}}{R}=\left(\frac{B L v}{R}\right) \cos \theta $$ in the loop. A wire carrying a current in a magnetic field experiences a force that is perpendicular to the plane defined by the wire and the magnetic field lines. The bar thus experiences a horizontal force \(\overrightarrow{\mathbf{F}}_{h}\) (perpendicular to the plane of \(\overrightarrow{\mathbf{B}}\) and the bar) given by $$ F_{h}=B I L=\left(\frac{B^{2} L^{2} v}{R}\right) \cos \theta $$ and shown in Fig However, we want the force component along the plane, which is $$ F_{\text {up plane }}=F_{h} \cos \theta=\left(\frac{B^{2} L^{2} v}{R}\right) \cos ^{2} \theta $$ When the bar reaches its terminal velocity, this force equals the gravitational force down the plane. Therefore, $$ \left(\frac{B^{2} L^{2} v}{R}\right) \cos ^{2} \theta=m g \sin \theta $$ from which the terminal speed is $$ v=\left(\frac{R m g}{B^{2} L^{2}}\right)\left(\frac{\sin \theta}{\cos ^{2} \theta}\right) $$ Can you show that this answer is reasonable in the limiting cases \(\theta\) \(=0, B=0\), and \(\theta=90^{\circ}\), and for \(R\) very large or very small?

A \(5.0-\Omega\) coil, of 100 turns and diameter \(6.0 \mathrm{~cm}\), is placed between the poles of a magnet so that the magnetic flux is maximum through the coil's cross-sectional area. When the coil is suddenly removed from the field of the magnet, a charge of \(1.0\) \(\times 10^{-4} \mathrm{C}\) flows through a \(595-\Omega\) galvanometer connected to the coil. Compute \(B\) between the poles of the magnet. As the coil is removed, the flux changes from \(B A\), where \(A\) is the coil's cross-sectional area, to zero. Therefore, $$ |\varepsilon|=N\left|\frac{\Delta \Phi_{M}}{\Delta t}\right|=N \frac{B A}{\Delta t} $$ We are told that \(\Delta q=1.0 \times 10^{-4} \mathrm{C} .\) But, by Ohm's Law, $$ |\varepsilon|=I R=\frac{\Delta q}{\Delta t} R $$ where \(R=600 \Omega\) is the total resistance. If we now equate these two expressions for \(|\varepsilon|\) and solve for \(\mathrm{B}\), we find $$ B=\frac{R \Delta q}{N A}=\frac{(600 \Omega)\left(1.0 \times 10^{-4} \mathrm{C}\right)}{(100)\left(\pi \times 9.0 \times 10^{-4} \mathrm{~m}^{2}\right)}=0.21 \mathrm{~T} $$

A coil having 50 turns of wire is removed in \(0.020 \mathrm{~s}\) from between the poles of a magnet, where its area intercepted a flux of \(3.1 \times 10^{-}\) \({ }^{4} \mathrm{~Wb}\), to a place where the intercepted flux is \(0.10 \times 10^{-4}\). Determine the average emf induced in the coil. $$ |\mathscr{E}|=N\left|\frac{\Delta \Phi_{M}}{\Delta t}\right|=50 \frac{(3.1-0.10) \times 10^{-4} \mathrm{~Wb}}{0.020 \mathrm{~s}}=0.75 \mathrm{~V} $$

The quarter-circle loop shown in has an area of \(15 \mathrm{~cm}^{2}\). A constant magnetic field, \(\mathrm{B}=0.16 \mathrm{~T}\), pointing in the \(+\chi\) -direction, fills the space independent of the loop. Find the flux through the loop in each orientation shown. The magnetic flux is determined by the amount of \(\overrightarrow{\mathbf{B}}\) -field passing perpendicularly through the particular area, times that area. That is, \(\Phi_{M}=B_{\perp} \mathrm{A} .\) (a) \(\Phi_{M}=B_{\perp} A=B A=(0.16 \mathrm{~T})\left(15 \times 10^{-4} \mathrm{~m}^{2}\right)=2.4 \times 10^{-4} \mathrm{~Wb}\) (b) \(\Phi_{M}=\left(B \cos 20^{\circ}\right) A=\left(2.4 \times 10^{-4} \mathrm{~Wb}\right)\left(\cos 20^{\circ}\right)=2.3 \times 10^{-4} \mathrm{~Wb}\) (c) \(\Phi_{M}=\left(B \sin 20^{\circ}\right) A=\left(2.4 \times 10^{-4} \mathrm{~Wb}\right)\left(\sin 20^{\circ}\right)=8.2 \times 10^{-5} \mathrm{~Wb}\)
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