/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 A 200-turn solenoid having a len... [FREE SOLUTION] | 91Ó°ÊÓ

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A 200-turn solenoid having a length of \(25 \mathrm{~cm}\) and a diameter of \(10 \mathrm{~cm}\) carries a current of \(0.29\) A. Calculate the magnitude of the magnetic field \(\vec{B}\) inside the solenoid.

Short Answer

Expert verified
The magnetic field inside the solenoid is approximately \( 2.91 \times 10^{-4} \text{ T} \).

Step by step solution

01

Identify Known Values

We are given that the solenoid has 200 turns, a length of 25 cm (which we convert to meters as 0.25 m), and carries a current of 0.29 A. The solenoid's diameter is 10 cm, but this is not needed for calculating the magnetic field inside a solenoid.
02

Understand the Magnetic Field Formula for a Solenoid

The magnetic field inside an ideal solenoid is calculated using the formula \( B = \mu_0 \cdot n \cdot I \), where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \text{ Tm/A}) \), \( n \) is the number of turns per unit length, and \( I \) is the current in the solenoid.
03

Calculate Number of Turns per Unit Length

The number of turns per unit length \( n \) is given by \( n = \frac{N}{L} \), where \( N \) is the total number of turns and \( L \) is the length of the solenoid. Thus, \( n = \frac{200}{0.25} = 800 \text{ turns/m} \).
04

Calculate the Magnetic Field

Substitute the values into the magnetic field formula: \( B = (4\pi \times 10^{-7}) \cdot 800 \cdot 0.29 \). Calculate \( B \):\[B = (4\pi \times 10^{-7} \text{ Tm/A}) \cdot 800 \cdot 0.29 = 2.91 \times 10^{-4} \text{ T}\]
05

Interpret the Result

The calculated magnetic field inside the solenoid is approximately \( 2.91 \times 10^{-4} \text{ Tesla} \). This value indicates the strength of the magnetic field due to the current passing through the solenoid in its interior.

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Key Concepts

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

Magnetic Field Calculation
To determine the magnetic field inside a solenoid, we employ a straightforward formula. The formula is: \( B = \mu_0 \cdot n \cdot I \). Here, \( B \) represents the magnetic field strength. Calculating this involves several steps:
  • First, we calculate the number of turns per unit length, denoted by \( n \), which is found using \( n = \frac{N}{L} \), where \( N \) is the total number of turns and \( L \) is the length of the solenoid.
  • Then, we substitute \( \mu_0 \), \( n \), and the current \( I \) into the formula to find the magnetic field strength \( B \).
By substituting the given values, like we did in the exercise, you will find the resulting magnetic field magnitude. In most calculations, dimensions such as the diameter of the solenoid are not necessary to determine its internal magnetic field. Overall, this approach gives a precise measure of how strong the magnetic field is inside the solenoid.
Permeability of Free Space
Permeability of free space, denoted as \( \mu_0 \), is a fundamental physical constant. It defines how well a magnetic field can be supported in a vacuum. The value of \( \mu_0 \) is \( 4\pi \times 10^{-7} \text{ Tm/A} \). This measurement is integral to calculating magnetic fields. Understanding \( \mu_0 \) is essential because it serves as a scaling factor in the magnetic field formula. It essentially tells us how the magnetic field will behave in a given medium—the vacuum being a perfect baseline scenario. The value of \( \mu_0 \) makes up part of the constants that help to calculate or predict how strong the magnetic field will be when considering the current and the construction of the solenoid. Remember, this constant does not change unless we are dealing with magnetic materials, which introduce the concept of relative permeability. But for a vacuum or air, \( \mu_0 \) remains constant.
Solenoid Properties
A solenoid is a coil of wire, typically in cylindrical form, that generates a magnetic field when an electric current passes through it. Solenoids are widely used due to their capability to convert electrical energy into a controlled magnetic field. Some key properties of a solenoid to understand include:
  • Number of Turns: This refers to how many loops the coil makes. More turns in the solenoid generally mean a stronger magnetic field for the same current.
  • Length: The length of the solenoid is crucial because it impacts the calculations for the number of turns per unit length, which in turn affects the magnetic field.
  • Current: The current flowing through the solenoid directly influences the magnetic field's strength. Higher currents result in stronger magnetic fields.
The diameter of the solenoid often does not play a role in the magnetic field strength when dealing with an ideal solenoid, where the field is uniform throughout its length.

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Most popular questions from this chapter

A surveyor is using a magnetic compass \(6.1 \mathrm{~m}\) below a power line in which there is a steady current of 100 A. (a) What is the magnetic field at the site of the compass due to the power line? (b) Will this field interfere seriously with the compass reading? The horizontal component of Earth's magnetic field at the site is \(20 \mu \mathrm{T}\).

Two wires, both of length \(L\), are formed into a circle and a square, and each carries current \(i\). Show that the square produces a greater magnetic field at its center than the circle produces at its center.

A long, hollow, cylindrical conductor (with inner radius \(2.0\) \(\mathrm{mm}\) and outer radius \(4.0 \mathrm{~mm}\) ) carries a current of 24 A distributed uniformly across its cross section. A long thin wire that is coaxial with the cylinder carries a current of \(24 \mathrm{~A}\) in the opposite direction. What is the magnitude of the magnetic field (a) \(1.0 \mathrm{~mm}\), (b) \(3.0 \mathrm{~mm}\), and (c) \(5.0 \mathrm{~mm}\) from the central axis of the wire and cylinder?

A long wire is known to have a radius greater than \(4.0 \mathrm{~mm}\) and to carry a current that is uniformly distributed over its cross section. The magnitude of the magnetic field due to that current is \(0.28 \mathrm{mT}\) at a point \(4.0 \mathrm{~mm}\) from the axis of the wire, and \(0.20 \mathrm{mT}\) at a point 10 \(\mathrm{mm}\) from the axis of the wire. What is the radius of the wire?

Equation \(29-4\) gives the magnitude \(B\) of the magnetic field set up by a current in an infinitely long straight wire, at a point \(P\) at perpendicular distance \(R\) from the wire. Suppose that point \(P\) is actually at perpendicular distance \(R\) from the midpoint of a wire with a finite length \(L\). Using Eq. \(29-4\) to calculate \(B\) then results in a certain percentage error. What value must the ratio \(L / R\) exceed if the percentage error is to be less than \(1.00 \%\) ? That is, what \(L / R\) gives

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