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Chapter 21: Problem 57

A piece of copper wire has a resistance per unit length of \(5.90 \times 10^{-3} \Omega / \mathrm{m}\). The wire is wound into a thin, flat coil of many turns that has a radius of \(0.140 \mathrm{~m}\). The ends of the wire are connected to a \(12.0-\mathrm{V}\) battery. Find the magnetic field strength at the center of the coil.

Short Answer

Expert verified
The magnetic field strength is calculated using the coil's geometry and current, yielding a field dependent on these parameters and constants.

Step by step solution

01

Calculate Total Length of the Wire

The resistance of a wire is given by the formula \( R = \rho \frac{L}{A} \), where \( L \) is the length of the wire, \( \rho \) is the resistivity, and \( A \) is the cross-sectional area. However, since resistance per unit length is provided as \( 5.90 \times 10^{-3} \Omega / \mathrm{m}\), we convert the resistance per unit length into total resistance: \( R = r \times L \), where \( r = 5.90 \times 10^{-3} \Omega/\mathrm{m} \). We are given the voltage \( V = 12.0 \mathrm{~V} \). Using Ohm's Law, \( R = \frac{V}{I} \), where \( I \) is the current.
02

Calculate the Current

Calculate the current using Ohm's Law. We need to express the total resistance in terms of the current: \( I = \frac{V}{R} = \frac{V}{rL} \). The resistance of the wire as wound into the coil equals the total resistance it presents, thus \( I = \frac{12.0}{5.90\times10^{-3} L} \) which simplifies into identifying \( I \) once \( L \) is known.
03

Express Length in Terms of Turns

The total length \( L \) of the wire is the circumference of one turn times the number of turns \( L = 2\pi RN \). Given \( N \) is the number of turns and \( R = 0.140 \mathrm{~m} \). Substitute \( L = N \times (2\pi \times 0.140) \) into the current equation from Step 2.
04

Magnetic Field of a Coil

The magnetic field \( B \) at the center of a coil of wire is given by the formula \( B = \frac{\mu_0 NI}{2R} \), where \( \mu_0 = 4\pi \times 10^{-7} \mathrm{~T~m/A} \) is the permeability of free space. Substitute the values for \( N \) and \( I \) obtained from previous calculations.
05

Solve for Magnetic Field Strength

Combine the above equations by substituting back \( N \) and \( L = N \times (2\pi \times 0.140) \) into \( B = \frac{\mu_0 N\frac{12.0}{0.1657 NL}}{2(0.140)} \) and simplify to find \( B \). This should cancel \( N \) out, leaving possibility to solve for \( B \) in terms of constants.

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Key Concepts

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

Resistance of a Wire
Understanding the resistance of a wire is crucial when analyzing circuits. The resistance of a wire depends on several factors:
  • The material of the wire, which is characterized by its resistivity (\( \rho \)). Copper, a common choice, has relatively low resistivity.
  • The length (\( L \)) of the wire; longer wires have higher resistance.
  • The cross-sectional area (\( A \)); as the area increases, resistance decreases.
When a wire is used to form a coil, the total resistance can be calculated by multiplying the resistance per unit length by the total length of the wire (\( R = r \times L \)). This allows us to understand how variations in these factors affect the overall resistance of the coil.
Ohm's Law
Ohm's Law is a fundamental principle useful in electricity and magnetism. It provides a simple relation between voltage (\( V \)), current (\( I \)), and resistance (\( R \)):\[V = IR\]This equation shows that the current flowing through a wire is directly proportional to the voltage applied and inversely proportional to the resistance. Thus, for a constant voltage, increasing resistance reduces the current, whereas reducing resistance increases it. Ohm's Law is essential in calculating the current that will flow in the circuit, especially when the resistance and voltage are known.
Current Calculation
To determine the current flowing through a coil, we rely on both resistance and voltage. Using Ohm's Law, current can be calculated as follows:\[I = \frac{V}{R}\]When you have a coil, like in this exercise, the resistance is dependent on the length of the wire used. The equation becomes:\[I = \frac{V}{rL}\]Where \( r \) is the resistance per unit length. By substituting the values for voltage and resistance per unit length into the equation, students can calculate the current. This calculation is critical to understanding how much magnetic field the coil will generate.
Magnetic Field Strength
The magnetic field strength at the center of a coil can be calculated using a well-known formula in electromagnetism:\[B = \frac{\mu_0 NI}{2R}\]This formula takes into account:
  • The permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \mathrm{~T~m/A} \)), which is a constant that relates magnetic fields to the factors creating them.
  • The number of turns (\( N \)) in the coil, as more turns increase the magnetic field strength.
  • The radius (\( R \)) of the coil, which inversely affects the strength鈥攍arger radii reduce field strength.
  • The current (\( I \)), calculated previously, directly affects the magnetic field.
By integrating all these factors, the magnetic field strength (\( B \)) can be calculated, offering insights into the behavior of electromagnets and the impact of different parameters.

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Most popular questions from this chapter

Two long, straight wires are separated by \(0.120 \mathrm{~m}\). The wires carry currents of \(8.0 \mathrm{~A}\) in opposite directions, as the drawing indicates. Find the magnitude of the net magnetic field at the points labeled (a) \(A\) and (b) \(B\).

A copper rod of length \(0.85 \mathrm{~m}\) is lying on a frictionless table (see the drawing). Each end of the rod is attached to a fixed wire by an unstretched spring that has a spring constant of \(k=75 \mathrm{~N} / \mathrm{m}\). A magnetic field with a strength of \(0.16 \mathrm{~T}\) is oriented perpendicular to the surface of the table. (a) What must be the direction of the current in the copper rod that causes the springs to stretch? (b) If the current is \(12 \mathrm{~A}\), by how much does each spring stretch?

Two circular loops of wire, each containing a single turn, have the same radius of \(4.0 \mathrm{~cm}\) and a common center. The planes of the loops are perpendicular. Each carries a current of 1.7 A. What is the magnitude of the net magnetic field at the common center?

A proton is projected perpendicularly into a magnetic field that has a magnitude of \(0.50 \mathrm{~T}\). The field is then adjusted so that an electron will follow the exact same circular path when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron? Verify that your answer is consistent with your answers to the Concept Questions.

A charge is moving perpendicular to a magnetic field and experiences a force whose magnitude is \(2.7 \times 10^{-3} \mathrm{~N}\). If this same charge were to move at the same speed and the angle between its velocity and the same magnetic field were \(38^{\circ},\) what would be the magnitude of the magnetic force that the charge would experience?
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