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Chapter 29: Problem 7

(II) A 16 -cm-diameter circular loop of wire is placed in a \(0.50-\mathrm{T}\) magnetic field. (a) When the plane of the loop is perpendicular to the field lines, what is the magnetic flux through the loop? (b) The plane of the loop is rotated until it makes a \(35^{\circ}\) angle with the field lines. What is the angle \(\theta\) in Eq. 1a for this situation? (c) What is the magnetic flux through the loop at this angle? \(\Phi_{B}=B_{\perp} A=B A \cos \theta=\vec{\mathbf{B}} \cdot \vec{\mathbf{A}} \quad[\vec{\mathbf{B}}\) uniform \(]\)

Short Answer

Expert verified
(a) 0.01005 T鈰卪虏; (b) \(\theta = 35^{\circ}\); (c) 0.00824 T鈰卪虏.

Step by step solution

01

Convert diameter to radius

The diameter of the loop is 16 cm; therefore, the radius is half of that. Convert the radius from centimeters to meters by dividing by 100.\( r = \frac{16}{2} = 8 \text{ cm} = 0.08 \text{ m} \)
02

Calculate the area of the loop

Use the formula for the area of a circle, \( A = \pi r^2 \), to calculate the area of the loop.\( A = \pi (0.08)^2 = 0.0201 \text{ m}^2 \)
03

Calculate magnetic flux when perpendicular

When the plane of the loop is perpendicular to the field lines, the angle \( \theta \) is 0. The formula for magnetic flux is \( \Phi_B = B A \cos \theta \). Calculate with \( \theta = 0 \).\( \Phi_B = 0.50 \times 0.0201 \times \cos(0^{\circ}) = 0.01005 \text{ T}\cdot\text{m}^2 \)
04

Identify the angle in question b

The loop is rotated to make a 35掳 angle with the field lines. The angle \( \theta \) in the context of the magnetic flux formula is the angle between the magnetic field and the normal to the loop. Thus, \( \theta = 35^{\circ} \).
05

Calculate magnetic flux at 35掳 angle

With \( \theta = 35^{\circ} \), plug the values into the magnetic flux formula to find the flux through the loop.\( \Phi_B = 0.50 \times 0.0201 \times \cos(35^{\circ}) = 0.00824 \text{ T} \cdot \text{m}^2 \)

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Key Concepts

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

Magnetic Field
A magnetic field is a vector field surrounding magnetic materials and electric currents. It represents the influence a magnetic source has over charged particles, causing them to experience a force. This force affects the path of moving charges or current-carrying conductors. Three important characteristics define a magnetic field:
  • Strength: Measured in teslas (T), it describes how robust the magnetic field is.
  • Direction: Since it is a vector, direction indicates the path along which magnetic force acts.
  • Uniformity: A uniform magnetic field has constant strength and direction across a region.
In our exercise, the magnetic field strength is given as 0.50 T. This field interacts with the circular loop to generate magnetic flux, which we'll explain further.
Circular Loop
A circular loop of wire, such as in the exercise, is often used in physics to explore magnetic effects. The loop acts as a closed path through which current runs or potential differences are explored. Several properties make circular loops significant in the context of magnetic fields:
  • Shape and Geometry: The circle extremity brings symmetry making calculations more straightforward.
  • Area: For magnetic flux considerations, the loop's area is vital. Calculated using the formula, \( A = \pi r^2 \), it refers to the surface through which the magnetic flux passes.
  • Induced Currents: When placed in a changing magnetic field, loops can have currents induced through them thanks to Faraday's law of induction.
In our problem, the diameter provided is 16 cm (0.16 m), making the radius 8 cm (0.08 m). This gives the loop an area of 0.0201 m虏.
Angle of Orientation
The angle of orientation is crucial in determining the magnetic flux through the loop. This angle, denoted as \( \theta \), is the angle between the magnetic field and the normal to the plane of the loop. Understanding this angle is important because:
  • Flux Maximization: Magnetic flux is at its maximum when the loop is perpendicular to the magnetic field lines (\( \theta = 0掳 \)), as cos(0掳) = 1.
  • Flux Minimization: Conversely, when the loop is parallel to the field lines (\( \theta = 90掳 \)), the flux is zero, since cos(90掳) = 0.
  • Variability of Flux: As the loop rotates, different amounts of the field penetrate it, depending on \( \theta \).
In our exercise, initially the loop is perpendicular for part (a) but in part (c), the angle is 35掳. Using cos(35掳) results in a decreased magnetic flux of 0.00824 T路m虏.

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Most popular questions from this chapter

(II) A 10.8 -cm-diameter wire coil is initially oriented so that its plane is perpendicular to a magnetic field of \(0.68 \mathrm{~T}\) pointing up. During the course of \(0.16 \mathrm{~s}\), the field is changed to one of \(0.25 \mathrm{~T}\) pointing down. What is the average induced emf in the coil?

(I) A simple generator is used to generate a peak output voltage of 24.0 \(\mathrm{V}\) . The square armature consists of windings that are 5.15 \(\mathrm{cm}\) on a side and rotates in a field of 0.420 \(\mathrm{T}\) at a rate of 60.0 \(\mathrm{rev} / \mathrm{s} .\) How many loops of wire should be wound on the square armature?

(II) The magnetic field perpendicular to a circular wire loop \(8.0 \mathrm{~cm}\) in diameter is changed from \(+0.52 \mathrm{~T}\) to \(-0.45 \mathrm{~T}\) in \(180 \mathrm{~ms},\) where \(+\) means the field points away from an observer and \(-\) toward the observer. \((a)\) Calculate the induced emf. (b) In what direction does the induced current flow?

(II) The back emf in a motor is 85 \(\mathrm{V}\) when the motor is operating at 1100 \(\mathrm{rpm} .\) How would you change the motor's magnetic field if you wanted to reduce the back emf to 75 \(\mathrm{V}\) when the motor was running at 2300 \(\mathrm{rpm}\) ?

(II) A dc generator is rated at \(16 \mathrm{~kW}, 250 \mathrm{~V}\), and \(64 \mathrm{~A}\) when it rotates at 1000 rpm. The resistance of the armature windings is \(0.40 \Omega\). ( \(a\) ) Calculate the "no-load" voltage at 1000 rpm (when there is no circuit hooked up to the generator). ( \(b\) ) Calculate the full-load voltage (i.e. at 64 A) when the generator is run at 750 rpm. Assume that the magnitude of the magnetic field remains constant.
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