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A solenoid is designed to produce a magnetic field of \(0.0270 \mathrm{~T}\) at its center. It has radius \(1.40 \mathrm{~cm}\) and length \(40.0 \mathrm{~cm},\) and the wire can carry a maximum current of 12.0 A. (a) What minimum number of turns per unit length must the solenoid have? (b) What total length of wire is required?

Short Answer

Expert verified
The minimum number of turns per unit length the solenoid must have is approximately \( 538605 \) turns per meter. The total length of wire required is approximately \( 150.77 \) meters.

Step by step solution

01

Calculate the number of turns per unit length

Firstly, we need to rearrange the formula for the magnetic field to solve for \( n = \frac{B}{\mu_0 * I} \). Here \( B = 0.0270 \mathrm{~T} \), \( \mu_0 = 4\pi * 10^{-7} \mathrm{Tm/A} \), and \( I = 12.0 \mathrm{A} \). By substituting these values into the rearranged formula, we find \( n = \frac{0.0270}{4\pi * 10^{-7} * 12.0} \) turns per meter.
02

Calculate the total length of wire

The total length of wire \( L \) equals the number of turns \( n \) times the circumference of the solenoid \( 2\pi r\), where \( r = 1.40 \mathrm{~cm} = 0.014 \mathrm{~m} \). So, \( L = n * 2\pi r \). Substituting for \( n \) from step 1, we find \( L = \frac{0.0270}{4\pi * 10^{-7} * 12.0} * 2\pi * 0.014 \) m.

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Key Concepts

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

Magnetic Field Strength
Understanding magnetic field strength is essential when dealing with electromagnetism. The magnetic field strength within a solenoid is a measure of the magnetizing force produced by an electric current flowing through the wire. This strength is directly proportional to the current, as well as the number of turns of wire per unit length of the solenoid. A key formula used to calculate this in a solenoid is given by: ewline ewline This mathematical representation elucidates that to achieve a desired magnetic field strength, one can adjust either the current through the wire or the density of the coil winding, measured in turns of wire per unit length. It is a pivotal concept for students to grasp, as it directly impacts the design and functionality of electromagnetic devices.
Turns Per Unit Length
The number of turns per unit length, often denoted by 'n', represents how tightly the wire is wound to form the solenoid. This density of turns is a critical parameter that has a direct impact on the magnetic field strength. The greater the number of turns per unit length, the stronger the magnetic field at the center of the solenoid, given a constant current. ewline ewline An increased number of turns per unit length indicates that more wire loops are packed into a given length, which enhances the magnetic field due to the additive effect of each loop's magnetic contribution. This concept underpins the calculations needed for designing a solenoid with particular magnetic properties and is essential for students to understand when determining the specifications for creating a strong magnetic field within a solenoid.
Solenoid Dimensions
The dimensions of a solenoid, which typically include its length and radius, play a vital role in determining its magnetic field strength and the amount of wire needed. The solenoid's length should be much greater than its radius to ensure a uniform magnetic field within its core, particularly near the center. ewline ewline A longer solenoid with a small radius maximizes the uniformity of the magnetic field inside. When calculating the total length of wire required to create a solenoid, dimensions are essential. The exercise demonstrates how the solenoid's circumference, which is derived from the radius, impacts the total wire length needed for winding. Students should appreciate how solenoid dimensions dictate both the physical construction and the resulting electromagnetic characteristics of the device.
Wire Current Capacity
Wire current capacity refers to the maximum electric current that a wire can safely carry without overheating and potentially causing damage or failure. This is a crucial safety and design consideration when creating a solenoid. ewline ewline The current passing through the solenoid's wire not only influences the magnetic field strength—as per the formula—but also must be maintained within the wire's tolerable limits. If a wire carries a current exceeding its capacity, it might lead to excessive heating, wire insulation damage, or even create a fire hazard. Hence, when solving for the solenoid design, as in the referenced problem, it is vital for students to understand that they must consider the wire's current capacity to ensure a safe and effective operating solenoid.

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Most popular questions from this chapter

Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current \(150 \mathrm{~A}\) and a height of \(8.0 \mathrm{~m}\) above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas and as a percentage of the earth's magnetic field, which is \(0.50 \mathrm{G}\). Is this value cause for worry?

A closely wound coil has a radius of \(6.00 \mathrm{~cm}\) and carries a current of 2.50 A. How many turns must it have if, at a point on the coil axis \(6.00 \mathrm{~cm}\) from the center of the coil, the magnetic field is \(6.39 \times 10^{-4} \mathrm{~T} ?\)

A long, straight wire lies along the \(y\) -axis and carries a current \(I=8.00 \mathrm{~A}\) in the \(-y\) -direction (Fig. E28.19). In addition to the magnetic field due to the current in the wire, a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) with magnitude \(1.50 \times 10^{-6} \mathrm{~T}\) is in the \(+x\) -direction. What is the total field (magnitude and direction) at the following points in the \(x z\) -plane: (a) \(x=0, z=1.00 \mathrm{~m} ;\) (b) \(x=1.00 \mathrm{~m}\) \(z=0 ;(\mathrm{c}) x=0, z=-0.25 \mathrm{~m} ?\)

A long, straight wire lies along the \(z\) -axis and carries a 4.00 A current in the \(+z\) -direction. Find the magnetic field (magnitude and direction) produced at the following points by a \(0.500 \mathrm{~mm}\) segment of the wire centered at the origin: (a) \(x=2.00 \mathrm{~m}, y=0, z=0\) (b) \(x=0, y=2.00 \mathrm{~m}, z=0\) (c) \(x=2.00 \mathrm{~m}, y=2.00 \mathrm{~m}, z=0\) (d) \(x=0, y=0, z=2.00 \mathrm{~m}\)

We can estimate the strength of the magnetic field of a refrigerator magnet in the following way: Imagine the magnet as a collection of current-loop magnetic dipoles. (a) Derive the force between two current loops with radius \(R\) and current \(I\) separated by distance \(d \ll R\). Very close to the wire its magnetic field is about the same as for an infinitely long wire, and Eq.( 28.11 ) can be used. (b) Using Eq. ( 28.17 ), express the current \(I\) in terms of the magnetic field at the middle of the loop, and express the radius \(R\) in terms of the area of the loop. In this way, derive an expression for the force \(F\) between two identical current loops separated by a small distance \(d\) in terms of their mutual area \(A\) and center magnetic field \(B\). (c) Rearrange your result to obtain an expression for the magnetic field of a dipole with area \(A\) in terms of the force \(F\) from an identical dipole separated by a small distance \(d\). (d) Now notice that the force it takes to separate one magnet from your refrigerator is nearly the same as the force it takes to separate two magnets stuck together. Estimate that force \(F\). (e) Estimate the area of a refrigerator magnet. (f) Assume that when these magnets are stuck together or to the refrigerator, they are separated by an effective distance \(d=25 \mu \mathrm{m}\). Use the formula derived above to estimate the magnetic field strength of the magnet.

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