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Chapter 22: Problem 16

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is \(7.0 \times 10^{-3}\) Wb. What is the flux that passes through the circular loop?

Short Answer

Expert verified
The magnetic flux through the circular loop is approximately 8.9 x 10^-3 Wb.

Step by step solution

01

Understand Magnetic Flux

Magnetic flux (\(\Phi\)) through a loop is defined as the product of the magnetic field (\(B\)) and the area (\(A\)) through which the field lines pass, perpendicular to both the field and the surface of the area. It can be expressed as \(\Phi = B \cdot A\).
02

Recall Area Formulas

The area of a square with side length \(s\) is calculated as \(s^2\), while the area of a circle with radius \(r\) is given by \(\pi r^2\). Since both loops are made from the same length of wire, we equate the perimeter of the square and the circumference of the circle.
03

Calculate Side Length of Square

Knowing that flux through the square loop is \(7.0 \times 10^{-3}\) Wb, and the formula \(\Phi = B \cdot s^2\), where \(s\) is the side of the square. The perimeter of the square is \(4s\).
04

Calculate Radius of Circle

Using the same wire length, the circumference of the circle is equal to the perimeter of the square, \(4s\). This gives the circumference as \(2\pi r = 4s\), which simplifies to \(r = \frac{2s}{\pi}\). Then, find the area \(A = \pi r^2 = \pi \left(\frac{2s}{\pi}\right)^2 = \frac{4s^2}{\pi}\).
05

Calculate Magnetic Flux Through the Circle

Because the magnetic field is unchanged, we express the flux through the circular loop as \(\Phi_{circle} = B \cdot \frac{4s^2}{\pi}\). Substitute the initial magnetic flux for \(B \cdot s^2\) in the formula, giving us \(\Phi_{circle} = \frac{4}{\pi} \Phi_{square}\).
06

Calculate Flux Through Circular Loop

Substitute \(\Phi_{square} = 7.0 \times 10^{-3}\) Wb into the equation to find the flux through the circular loop: \(\Phi_{circle} = \frac{4}{\pi} \times 7.0 \times 10^{-3}\) Wb, which simplifies to approximately \(8.9 \times 10^{-3}\) Wb.

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Key Concepts

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

Magnetic Field
The magnetic field is a fundamental concept in physics. It represents a region around a magnetic material or moving electric charge where forces of magnetism act.
Magnetic fields are often depicted using lines of force that show the direction and strength of the field. These lines are denser where the magnetic field is stronger.
  • Magnetic fields are created by magnets or moving charges.
  • The field's strength and direction can influence other magnets or charged particles within its reach.
  • Field lines enter the south pole and exit the north pole of a magnet.
These fields play a vital role in many technological and natural phenomena, affecting everything from electric motors to the Earth's own geomagnetic field.
Square Loop
A square loop is simply a loop of wire shaped as a square. When placed in a magnetic field, it interacts with the field lines.
This interaction can help us understand electromagnetic effects such as magnetic flux. The side length of the square determines its area, calculated as: \[ A_{square} = s^2 \] where \(s\) is the length of one side of the square.
  • The total length of wire forms the perimeter of the square, calculated by \(4s\).
  • When perpendicular to the magnetic field, the square loop captures maximum flux.
In practical applications, square loops are used in devices like transformers and sensors, where precise geometry affects functionality.
Circular Loop
When the same wire is reshaped into a circular loop, the geometry changes, but the wire's length remains the same. This transformation gives rise to different calculations for area and flux.
  • The circumference of the circle is equivalent to the perimeter of the square: \(2\pi r = 4s\).
  • From this, we find the radius: \(r = \frac{2s}{\pi}\).
The area of a circular loop is:\[ A_{circle} = \pi r^2 = \frac{4s^2}{\pi} \] This new area influences the magnetic flux that passes through the loop, illustrating how shape impacts physical properties.
Magnetic Flux Calculation
Magnetic flux, denoted by \( \Phi \), quantifies the total magnetic field passing through a given area. It combines both the field strength and the area through which it acts.
  • The basic formula for magnetic flux is: \( \Phi = B \cdot A \), where \(B\) is the magnetic field strength.
  • For the given problem, the flux through a square loop is known: \(7.0 \times 10^{-3}\) Wb.
For the circular loop, we use the relationship:\[ \Phi_{circle} = \frac{4}{\pi} \Phi_{square} \] This equation stems from the change in area due to the new circular shape, leading to a calculated flux of approximately \(8.9 \times 10^{-3}\) Wb. This exercise highlights how the same physical material can produce different effects based on its shape and orientation.

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Most popular questions from this chapter

In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil \(1,\) which has 184 loops, is \(2.82 \mathrm{V}\). The emf induced in coil 2 is 4.23 V. How many loops does coil 2 have?

A \(5.40 \times 10^{-5} \mathrm{H}\) solenoid is constructed by wrapping 65 turns of wire around a cylinder with a cross-sectional area of \(9.0 \times 10^{-4} \mathrm{m}^{2} .\) When the solenoid is shortened by squeezing the turns closer together, the inductance increases to \(8.60 \times 10^{-5}\) H. Determine the change in the length of the solenoid.

A \(0.80-\mathrm{m}\) aluminum bar is held with its length parallel to the east- west direction and dropped from a bridge. Just before the bar hits the river below, its speed is \(22 \mathrm{m} / \mathrm{s},\) and the emf induced across its length is \(6.5 \times 10^{-4} \mathrm{V} .\) Assuming the horizontal component of the earth's magnetic field at the location of the bar points directly north, (a) determine the magnitude of the horizontal component of the earth's magnetic field, and (b) state whether the east end or the west end of the bar is positive.

You need to design a \(60.0-\mathrm{Hz}\) ac generator that has a maximum emf of \(5500 \mathrm{V}\). The generator is to contain a 150 -turn coil that has an area per turn of \(0.85 \mathrm{m}^{2} .\) What should be the magnitude of the magnetic field in which the coil rotates?

A rectangular loop of wire with sides 0.20 and \(0.35 \mathrm{m}\) lies in a plane perpendicular to a constant magnetic field (see part \(a\) of the drawing). The magnetic field has a magnitude of \(0.65 \mathrm{T}\) and is directed parallel to the normal of the loop's surface. In a time of 0.18 s, one-half of the loop is then folded back onto the other half, as indicated in part \(b\) of the drawing. Determine the magnitude of the average emf induced in the loop.
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