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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 \(5.50 \times 10^{-3}\) Wb.

Step by step solution

01

Understand Magnetic Flux Formula

The magnetic flux \( \Phi \) through a loop is given by the formula \( \Phi = B \cdot A \cdot \cos(\theta) \), where \( B \) is the magnetic field strength, \( A \) is the area of the loop, and \( \theta \) is the angle between the field and the normal to the loop's surface. Since the loop is perpendicular to the magnetic field, \( \theta = 0^\circ \), and \( \cos(0^\circ) = 1 \). Therefore, \( \Phi = B \cdot A \).
02

Calculate Area of Square Loop

For the square loop, \( \Phi_{\text{square}} = B \times A_{\text{square}} \). Given that \( \Phi_{\text{square}} = 7.0 \times 10^{-3} \) Wb, we can express the area of the square as \( A_{\text{square}} = \frac{\Phi_{\text{square}}}{B} \).
03

Equate Magnetic Flux

The magnetic field \( B \) remains constant when the loop is re-formed into a circular loop. Therefore, the magnetic flux before and after reforming the loop will involve the same magnetic field, \( B \). This means the area of each shape must correspond to their respective flux when multiplied by \( B \).
04

Use Area Conversion for Circular Loop

Since the wire length does not change, the perimeter of the square equals the circumference of the circle. If the side length of the square is \( s \), then \( 4s = 2\pi r \), where \( r \) is the circle's radius. Using the formula for perimeter equality: \( s = \frac{\pi r}{2} \).
05

Derive Area Relationship

From \( A_{\text{square}} = s^2 \) and the relationship \( s = \frac{\pi r}{2} \), solve for \( r \). The area of the circle will be \( A_{\text{circle}} = \pi r^2 \). Since \( 4s = 2\pi r \), substituting for \( s \) in terms of \( r \), we solve for \( r \) and subsequently \( A_{\text{circle}} : A_{\text{square}} = \pi r^2 : s^2 \).
06

Solve for Flux Through Circle

This yields an inferred ratio \( A_{\text{circle}} = \frac{\pi}{4} A_{\text{square}} \). Therefore, \( \Phi_{\text{circle}} = B \times A_{\text{circle}} = \frac{\pi}{4} \times B \times A_{\text{square}} = \frac{\pi}{4} \times \Phi_{\text{square}} = \frac{\pi}{4} \times 7.0 \times 10^{-3} \).
07

Calculate Numerical Flux Value

Replacing \( \Phi_{\text{square}} \), the magnetic flux through the circular loop: \[ \Phi_{\text{circle}} = \frac{\pi}{4} \times 7.0 \times 10^{-3} \approx 5.50 \times 10^{-3} \text{ Wb} \].

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Key Concepts

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

Circular Loop
When we talk about a circular loop in the context of magnetic flux, it refers to a wire bent into the shape of a circle. The concept is crucial because changing shapes can impact how much magnetic flux passes through it.
For the magnetic flux to be calculated in such a circle, the loop's area is key because the larger the area within the loop, the more magnetic flux can pass through.
A unique property of the circular loop is its symmetrical shape, which often results in efficient flux flow when the magnetic field lines are perpendicular to the loop. Although the amount of wire remains unchanged, reforming it from a square to a circle changes the area, impacting the magnetic flux according to the formula, \( \Phi = B \cdot A \).
Square Loop
The square loop is essentially a loop formed into a square shape. In the exercise, this square loop is initially exposed to a uniform magnetic field.
The sides of the square define its perimeter, which directly affects the area over which the magnetic field flux is calculated.
  • A square shape inherently maximizes the magnetic flux within a given perimeter compared to other more complex shapes.
  • For instance, if the square has side length \( s \), its area is \( s^2 \).
Since it's perpendicular to the magnetic field, the entire area contributes to the flux calculation, making it \( \Phi = B \cdot s^2 \). This uniformity means the magnetic flux passing through remains straightforward to calculate.
Magnetic Field
The magnetic field is an invisible field that exerts force on substances which are magnetic or electric. For our consideration, it affects the calculated magnetic flux through the loop.
When evaluating magnetic flux, the strength and uniformity of the magnetic field鈥攄enoted as \( B \)鈥攑lays a significant role. This value, \( B \), remains constant regardless of the shape of the loop, whether it's a square or circular.
  • Magnetic field direction makes an impact since the loop is perpendicular, optimizing the flux passing through.
  • Changes in field strength can alter the amount of flux through a loop.
Because the field stays uniform throughout the process, it simplifies flux calculation using the loop area alone for both shapes.
Area Conversion
Area conversion is a critical part of understanding how reshaping a loop influences magnetic flux. The tricky part is to compute how a loop鈥檚 area changes when its shape changes but not its perimeter.
When converting a square loop into a circular loop, the area of the circle differs from that of the square even if the wire length鈥攐r perimeter鈥攕tays the same.
  • The square's perimeter \( 4s \) must equal the circle's circumference \( 2\pi r \).
  • This results in an inherent relation: \( s = \frac{\pi r}{2} \).
  • From this perimeter equality, the circular area \( A_{\text{circle}} \) is derived using \( \pi r^2 \).
This transformation affects the flux, with the circular loop exhibiting different magnetic properties due to this change in area, previously calculated as being \( \frac{\pi}{4} \) times the square's area.

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Most popular questions from this chapter

Suppose there are two transformers between your house and the high-voltage transmission line that distributes the power. In addition, assume that your house is the only one using electric power. At a substation the primary coil of a step-down transformer (turms ratio \(=1 : 29\) ) receives the voltage from the high-voltage transmission line. Because of your usage, a current of 48 mA exists in the primary coil of this transformer. The secondary coil is connected to the primary of another step-down transformer (turns ratio \(=1 : 32 )\) somewhere near your house, perhaps up on a telephone pole. The secondary coil of this transformer delivers a \(240-\mathrm{V}\) emf to your house. How much power is your house using? Remember that the current and voltage given in this problem are rms values.

In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the trans- former is connected to a \(120-\mathrm{V}\) receptacle on a wall. The picture tube of the television set uses 91 \(\mathrm{W}\) , and there is 5.5 \(\mathrm{mA}\) of current in the secondary coil of the transformer to which the tube is connected. Find the turns ratio \(N_{s} / N_{\mathrm{p}}\) of the transformer.

A solenoid has a cross-sectional area of \(6.0 \times 10^{-4} \mathrm{m}^{2},\) consists of 400 turns per meter, and carries a current of 0.40 \(\mathrm{A}\) . A 10 -turn coil is wrapped tightly around the circumference of the solenoid. The ends of the coil are connected to a \(1.5-\Omega\) resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time of 0.050 s. Find the average current induced in the coil.

The magnetic flux that passes through one turn of a 12 -turn coil of wire changes to 4.0 \(\mathrm{from} 9.0 \mathrm{Wb}\) in a time of 0.050 \(\mathrm{s}\) . The average induced current in the coil is 230 \(\mathrm{A}\) . What is the resistance of the wire?

*27. ssm A magnetic field is passing through a loop of wire whose area is 0.018 m2 The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s. (a) Determine the magnitude of the emf induced in the loop. (b) Suppose that the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m2 /s) should the area be changed at the instant when B 1.8 T if the induced emf is to be zero? Explain whether the area is to be enlarged or shrunk.
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