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(a) At what angle \(\theta\) is the torque on a current loop 90.0\% of maximum? (b) 50.0\% of maximum? (c) 10.0\% of maximum?

Short Answer

Expert verified
The angles at which the torque on a current loop is (a) 90.0% of maximum, (b) 50.0% of maximum, and (c) 10.0% of maximum are approximately (a) \(\theta \approx 64.16^\circ\), (b) \(\theta = 30^\circ\), and (c) \(\theta \approx 5.74^\circ\).

Step by step solution

01

Calculate 90% of maximum torque value

First, let's get the torque value when it's at 90% of the maximum torque: \[ \tau_{90\%} = 0.9 \times \tau_{max} = 0.9 \times nIA \]
02

Find the angle for 90% of maximum torque

Now, we can write the equation for the torque when it's at 90% of the maximum torque: \[ 0.9nIA = nIA \sin \theta \] Divide both sides by nIA: \[ 0.9 = \sin \theta \] Now, to find the angle, we'll take the inverse sine of both sides: \[ \theta = \sin^{-1}(0.9) \] Applying the inverse sine gives us the angle at which the torque is 90% of the maximum torque: \[ \theta \approx 64.16^\circ \]
03

Find the angle for 50% of maximum torque

We'll follow similar steps for the 50% case. Find the torque value when it's at 50% of the maximum torque: \[ \tau_{50\%} = 0.5 \times \tau_{max} = 0.5 \times nIA \] Now write the equation for the torque when it's at 50% of the maximum torque: \[ 0.5nIA = nIA \sin \theta \] Divide both sides by nIA: \[ 0.5 = \sin \theta \] Take the inverse sine of both sides to find the angle: \[ \theta = \sin^{-1}(0.5) \] This gives us the angle at which the torque is 50% of the maximum torque: \[ \theta = 30^\circ \]
04

Find the angle for 10% of maximum torque

Lastly, let's find the torque value when it's at 10% of maximum torque: \[ \tau_{10\%} = 0.1 \times \tau_{max} = 0.1 \times nIA \] Now write the equation for the torque when it's at 10% of the maximum torque: \[ 0.1nIA = nIA \sin \theta \] Divide both sides by nIA: \[ 0.1 = \sin \theta \] Take the inverse sine of both sides to find the angle: \[ \theta = \sin^{-1}(0.1) \] This gives us the angle at which the torque is 10% of the maximum torque: \[ \theta \approx 5.74^\circ \] So the angles are: a) 90% maximum torque: \(\theta \approx 64.16^\circ\) b) 50% maximum torque: \(\theta = 30^\circ\) c) 10% maximum torque: \(\theta \approx 5.74^\circ\)

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Key Concepts

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

Current Loop
A current loop is a closed circuit through which electric current flows, forming a loop-like structure, often rectangular or circular. This loop plays a crucial role in generating magnetic fields. When placed in a magnetic field, it experiences torque, which is the rotational force acting on the loop. This setup is fundamental in electromagnetism and forms the basis of many devices like motors and meters.

Key properties of a current loop include:
  • Current (I): The flow of electric charge through the loop.

  • Area (A): The surface area enclosed by the loop.

  • Number of turns (n): Indicates how many times the wire winds around in the loop, which enhances the magnetic effect.
Each of these factors contributes to the torque experienced by the loop when placed in a magnetic field, as seen with the equation: \[ \tau = nIA \sin \theta \]where \( \tau \) is the torque and \( \theta \) is the angle between the normal to the loop and the magnetic field.
Angle Calculation
When trying to find specific angles at which a current loop's torque reaches a certain percentage of its maximum, angle calculation comes into play. The torque is dependent on the sine of the angle \( \theta \). To find angles such as when the torque is 90%, 50% or 10% of the maximum, we rearrange the torque equation:

\[ \sin \theta = \frac{\text{desired torque}}{\text{maximum torque}} \]

For a current loop, the maximum torque occurs when \( \theta = 90^\circ \), since \( \sin 90^\circ = 1 \). Therefore, to find any particular angle \( \theta \):
  • Calculate the ratio of the specific torque you want to the maximum torque.

  • Use the sine equation to calculate \( \sin \theta \).

  • Solve for \( \theta \) using the inverse sine function.
By these steps, angles such as \( 64.16^\circ, 30^\circ, \) and \( 5.74^\circ \) are found, corresponding to 90%, 50%, and 10% of maximum torque levels, respectively.
Inverse Sine Function
The inverse sine function, denoted as \( \sin^{-1}(x) \) or arcsin\( (x) \), is crucial in finding angles when dealing with trigonometric equations. When you know the sine of an angle, this function helps you determine the actual angle itself.

Here's how you can use the inverse sine function efficiently:
  • Given \( \sin \theta = x \), find \( \theta \) by calculating \( \sin^{-1}(x) \).

  • Make sure \( x \) is within the interval \(-1 \leq x \leq 1\), since these are the bounds of the sine function.

  • The result \( \theta \) will be within the range \([-90^\circ, 90^\circ]\).

  • For angles outside this range, consider the periodic nature of the sine function or symmetry.
In the context of torque in a current loop, calculating \( \sin^{-1}(0.9) \), \( \sin^{-1}(0.5) \), and \( \sin^{-1}(0.1) \) helps us find angles where the torque reaches certain percentages of its maximum value, namely, around \( 64.16^\circ \), \( 30^\circ \), and \( 5.74^\circ \), respectively.

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Most popular questions from this chapter

(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a \(0.500-\mu C\) charge and flies due west at a speed of \(660 .\) m/s over Earth's south magnetic pole, where the \(8.00 \times 10^{-5}-\mathrm{T}\) magnetic field points straight down into the ground. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect.

What magnetic field is required in order to confine a proton moving with a speed of \(4.0 \times 10^{6} \mathrm{m} / \mathrm{s}\) to a circular orbit of radius \(10 \mathrm{cm} ?\)

(a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less than \(1.00 \times 10^{-12} \mathrm{N}\). What is the greatest the charge can be if it moves at a maximum speed of \(30.0 \mathrm{m} / \mathrm{s}\) in Earth's field? (b) Discuss whether it would be difficult to limit the charge to less than the value found in (a) by comparing it with typical static electricity and noting that static is often absent.

An electron in a TV CRT moves with a speed of \(6.0 \times 10^{7} \mathrm{m} / \mathrm{s}, \quad\) in a direction perpendicular to Earth's field, which has a strength of \(5.0 \times 10^{-5} \mathrm{T}\). (a) What strength electric field must be applied perpendicular to the Earth's field to make the electron moves in a straight line? (b) If this is done between plates separated by \(1.00 \mathrm{cm}\) what is the voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a correction.)

An electron and a proton move with the same speed in a plane perpendicular to a uniform magnetic field. Compare the radii and periods of their orbits.

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