/*! 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 52 The plane of a rectangular loop ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 52

The plane of a rectangular loop of wire with sides \(0.05\) and \(0.08 \mathrm{~m}\) is parallel to a uniform magnetic field of induction \(1.5\) \(\times 10^{-2} \mathrm{~T}\). A current of \(10.0\) ampere flows through the loop. If the side of length \(0.08 \mathrm{~m}\) is normal and the side of length \(0.05\) \(m\) is parallel to the lines of induction, then the torque acting on the loop is a. \(6000 \mathrm{Nm}\) b zero c. \(1.2 \times 10^{-2} \mathrm{Nm}\) d \(6 \times 10^{-4} \mathrm{Nm}\)

Short Answer

Expert verified
Indicating a clerical issue, the closest calculated torque is \(1.2 \times 10^{-4} \, \text{Nm}\), not matching any options directly.

Step by step solution

01

Identify the Torque Equation

The torque (蟿) on a current-carrying loop in a magnetic field can be calculated using the equation: \[蟿 = n imes I imes A imes B imes ext{sin}( heta),\]where \(n\) is the number of turns in the loop, \(I\) is the current, \(A\) is the area of the loop, \(B\) is the magnetic induction (magnetic field strength), and \(\theta\) is the angle between the normal to the loop and the magnetic field.
02

Calculate the Loop Area

The area \(A\) of the rectangular loop is found by multiplying its length and width. Here, the sides are \(0.05\) m and \(0.08\) m, thus the area is:\[A = 0.05 \, \text{m} \times 0.08 \, \text{m} = 0.004 \, \text{m}^2.\]
03

Determine the Angle 胃

Since the side of length \(0.08\) m is normal to the magnetic field, \(\theta = 90^{\circ}\). The sine of \(90^{\circ}\) is 1, so:\[\text{sin}(\theta) = \text{sin}(90^{\circ}) = 1.\]
04

Substitute Values into the Torque Equation

Substitute the known values into the torque equation. Here, \(n = 1\) (one loop), \(I = 10.0\) A, \(A = 0.004\) m虏, \(B = 1.5 \times 10^{-2}\) T, and \(\text{sin}(\theta) = 1\):\[蟿 = 1 \times 10.0 \, \text{A} \times 0.004 \, \text{m}^2 \times 1.5 \times 10^{-2} \, \text{T} \times 1.\]Solving this gives:\[蟿 = 1.2 \times 10^{-4} \, \text{Nm}.\]
05

Compare with Given Options

Upon rounding the calculated torque, compare it with the given options:a. \(6000 \, \text{Nm}\) b. zero c. \(1.2 \times 10^{-2} \, \text{Nm}\) d. \(6 \times 10^{-4} \, \text{Nm}\)None of these exactly match \(1.2 \times 10^{-4} \, \text{Nm}\), indicating a need to verify calculations or that there may be a misinterpretation in the setup/units in the options provided.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Magnetic Induction
Magnetic induction, also known as magnetic field strength or magnetic flux density, is a measure of the magnetic field's influence on its environment. It is denoted by the symbol \(B\) and is measured in teslas (T). This field affects any magnetic, current-carrying object within its vicinity. Magnetic induction becomes particularly relevant when considering torque on a current-carrying loop, as it is one of the factors in the formula for calculating torque. In this context, the uniform magnetic field has an induction value of \(1.5 \times 10^{-2} \, \mathrm{T}\).

Magnetic induction interacts with current-carrying conductors by exerting forces on the charges in motion. This is critical for the operation of devices such as electric motors and generators. Understanding magnetic induction involves recognizing it as part of a larger electromagnetic interaction, where magnetic fields exert influence directly correlated to their strength and configuration.
Rectangular Loop Area Calculation
Calculating the area of a rectangular loop is a straightforward process, essential for determining the torque induced on the loop by a magnetic field. The area \(A\) is calculated by multiplying the loop's length by its width.

For the problem at hand, the rectangular loop has lengths of \(0.05 \, \text{m}\) and \(0.08 \, \text{m}\). The calculation of the area is simple:
  • \(A = \text{length} \times \text{width} = 0.05 \, \text{m} \times 0.08 \, \text{m}\)
  • This results in an area of \(0.004 \, \text{m}^2\)
This area becomes a critical factor when inputting into the torque formula, as it directly influences the magnitude of the torque generated by the magnetic field applied to the current-carrying loop.
Angle Between Loop and Magnetic Field
The angle \(\theta\) between the loop and the magnetic field is a pivotal part of predicting the loop's behavior under the influence of the field. This angle is specifically the angle between the normal to the loop's plane and the direction of the magnetic field.

In this scenario, one side of the rectangle is described as normal (perpendicular) to the field. As such, \(\theta = 90^\circ\). At an angle of \(90^\circ\), the sine function, which is part of the torque calculation, reaches its maximum value of 1, simplifying the torque calculation to:
  • \(\text{sin}(\theta) = \text{sin}(90^\circ) = 1\)
When the magnetic field is perpendicular to the plane of the loop, the calculated torque will be at its maximum value, making \(\theta\) a critical parameter in determining the effectiveness of the magnetic field's impact.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An insulating rod of length \(\ell\) carries a charge \(q\) distributed uniformly on it. The rod is pivoted at its mid point and is rotated at a frequency \(f\) about a fixed axis perpendicular to the rod and passing through the pivot. The magnetic mornent of the rod system is a. \(\frac{1}{12} \pi q f \ell^{2}\) b \(\pi_{q} f \ell^{2}\) c. \(\frac{1}{6} \pi q f \ell^{2}\) d. \(\frac{1}{3} \pi q f \ell^{2}\)

A charged particle of specific charge (charge/mass) \(\alpha\) is released from origin at time \(t=0\) with velocity \(\vec{v}=v_{0}(i+\hat{j})\) in uniform magnetic field \(\bar{B}=B_{0} \hat{i}\). Coordinates of the particle at time \(t=\frac{\pi}{B_{0} \alpha}\) are \(\mathrm{a}\left(\frac{v_{0}}{2 B_{0} \alpha}, \frac{\sqrt{2} v_{0}}{\alpha B_{0}} ; \frac{-v_{0}}{B_{0} \alpha}\right)\) b \(\left(\frac{-v_{0}}{2 B_{0} \alpha}, 0 ; 0\right)\) c. \(\left(0, \frac{2 v_{0}}{B_{0} \alpha}, \frac{v_{0} \pi}{2 B_{0} \alpha}\right)\) \(d\left(\frac{v_{0} \pi}{B_{0} \alpha}, 0, \frac{-2 v_{0}}{B_{0} \alpha}\right)\)

The magnetic field due to a current carrying circular loop of radius \(3 \mathrm{~cm}\) at a point on the axis at a distance of \(4 \mathrm{~cm}\) from the center is \(54 \mathrm{mT}\). Its value at the center of the loop will be a \(250 \mu \mathrm{T}\) b \(150 \mu \mathrm{T}\) c. \(125 \mu \mathrm{T}\) d. \(75 \mu \mathrm{T}\)

A current \(l\) flows a thin wire shaped as regular polygon of \(n\) sides which can be inscribed in a circle of radius \(R\). The magnetic field induction at the center of polygon due to one side of the polygon is a \(\frac{\mu_{0} I}{\pi R}\left(\tan \frac{\pi}{n}\right)\) \& \(\frac{\mu_{0} I}{4 \pi R} \tan \frac{\pi}{n}\) c. \(\frac{\mu_{0} I}{2 \pi R}\left(\tan \frac{\pi}{n}\right)\) d \(\frac{\mu_{0} I}{2 \pi R}\left(\cos \frac{\pi}{n}\right)\)

A particle of charge \(-1.6 \times 10^{-18} \mathrm{C}\) moving with velocity 10 \(\mathrm{ms}^{-1}\) along the \(x\) -axis enters a region where a magnetic field of induction \(B\) is along the \(y\) -axis, and an electric ficld of magnitude \(10 \mathrm{Vm}^{-1}\) is along the negative \(z\) -axis. If the charged particle c枚ntinues moving along the \(x\) -axis, the magnitude of \(B\) is a. \(10^{-3} \mathrm{Wbm}^{-2}\) b \(10^{3} \mathrm{Wbm}^{-2}\) c. \(10^{2} \mathrm{Wbm}^{-2}\) d. \(10^{16} \mathrm{Wbm}^{-2}\)
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Work Energy and Power

Read Explanation

Quantum Physics

Read Explanation

Turning Points in Physics

Read Explanation

Magnetism

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.