/*! 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 13 ssm A standard door into a house... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 13

ssm A standard door into a house rotates about a vertical axis through one side, as defined by the door's hinges. A uniform magnetic field is parallel to the ground and perpendicular to this axis. Through what angle must the door rotate so that the magnetic flux that passes through it decreases from its maximum value to one-third of its maximum value?

Short Answer

Expert verified
The door must rotate approximately 70.53 degrees.

Step by step solution

01

Understanding Magnetic Flux Formula

Magnetic flux (桅) through a surface is given by the formula: \[ \Phi = B \cdot A \cdot \cos(\theta) \] where B is the magnetic field, A is the area of the surface, and \( \theta \) is the angle between the magnetic field and the normal (perpendicular) to the surface plane.
02

Analyze Initial Conditions

Initially, when the magnetic flux is at its maximum, the angle \( \theta \) is 0 degrees (meaning the door's surface is perpendicular to the magnetic field lines). Thus, the maximum flux is \[ \Phi_{\text{max}} = B \cdot A \cdot \cos(0) = B \cdot A \].
03

Calculate One-third of Maximum Flux

We want the magnetic flux to decrease to one-third of its maximum value: \[ \Phi = \frac{1}{3} \Phi_{\text{max}} = \frac{1}{3} (B \cdot A) \].
04

Set Up Equation for One-third Flux

Using the formula for magnetic flux, set up the equation when the flux is one-third of the maximum value:\[ B \cdot A \cdot \cos(\theta) = \frac{1}{3} (B \cdot A) \]. The terms \( B \cdot A \) cancel out on both sides, simplifying to:\[ \cos(\theta) = \frac{1}{3} \].
05

Calculate the Angle

We solve for \( \theta \) using the relation \[ \cos(\theta) = \frac{1}{3} \]. Using the inverse cosine function, we find:\[ \theta = \cos^{-1}(\frac{1}{3}) \approx 70.53^{\circ} \].

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 Field
A magnetic field is a fundamental concept in physics that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. Magnetic fields are usually represented by the letter 'B'. They are invisible forces that can exert magnetic forces on other objects within their vicinity.

When we talk about the magnetic field in the context of a rotating door problem, we're interested in how this field interacts with the surface of the door. Here鈥檚 what you should keep in mind:
  • **Direction**: The magnetic field is parallel to the ground and perpendicular to the door鈥檚 rotation axis.
  • **Interaction with surfaces**: If a surface is perpendicular to the field lines, the magnetic flux through it is maximized, as in the initial state of the door in the given problem.
Understanding the orientation and strength of the magnetic field is crucial when calculating how much of that field interacts with a surface like our door.
Angle of Rotation
The angle of rotation is central to understanding how the magnetic flux through a surface changes. When dealing with a door or similar surfaces rotating through a magnetic field:
  • **Initial position**: Initially, when the door is perpendicular to the magnetic field, the angle, denoted by \( \theta \), is zero degrees.
  • **Change in angle**: As the door rotates from this position, the angle \( \theta \) increases, consequently reducing the magnetic flux.
For our door exercise, solving the problem requires calculating the specific angle of rotation where the magnetic flux becomes one-third of its maximum value. The problem is resolved by applying trigonometric concepts to find that rotated angle (\( 70.53^{\circ} \)), as discovered through the cosine function formula.
Cosine Function
The cosine function, often represented as \( \cos(\theta) \), measures the cosine of the angle \( \theta \) between two vectors. In physics, it鈥檚 widely used to correlate various components of vectors, especially in problems involving forces and fields.

In our exercise, the cosine function helps determine the relationship between the angle of the door and the magnetic flux passing through it.
  • **Flux Relation**: As the door rotates, the angle \( \theta \) changes, and we use \( \cos(\theta) \) to calculate how much of the magnetic field interacts with the door's surface.
  • **Rotation Impact**: A smaller \( \cos(\theta) \) implies a larger angle of rotation, leading to lesser magnetic flux.
Using the formula \( \cos(\theta) = \frac{1}{3} \), we solve it to find \( \theta = \cos^{-1}(\frac{1}{3}) \), approximately equal to \( 70.53^{\circ} \), thereby identifying the necessary rotation angle to achieve one-third of the maximum magnetic flux.

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

A 120.0 -V motor draws a current of 7.00 \(\mathrm{A}\) when running at normal speed. The resistance of the armature wire is 0.720\(\Omega .\) (a) Determine the back emf generated by the motor. (b) What is the current at the instant when the motor is just turned on and has not begun to rotate? (c) What series resistance must be added to limit the starting current to 15.0 \(\mathrm{A}\) ?

A generator uses a coil that has 100 turns and a \(0.50-\) T magnetic field. The frequency of this generator is 60.0 \(\mathrm{Hz}\) , and its emf has an rms value of 120 \(\mathrm{V}\) . Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made.

ssm A circular coil \((950 \text { turns, radius }=0.060 \mathrm{m})\) is rotating in a uniform magnetic field. At \(t=0\) s, the normal to the coil is perpendicular to the magnetic field. At \(t=0.010\) s, the normal makes an angle of \(\phi=45^{\circ}\) with the field because the coil has made one-eighth of a revolution. An average emf of magnitude 0.065 \(\mathrm{V}\) is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

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 generator is connected across the primary coil ( \(N_{p}\), turns) of a transformer, while a resistance \(R_{2}\) is connected across the secondary coil \(\left(N_{\mathrm{s}}\right.\) turns). This circuit is equivalent to a circuit in which a single resistance \(R_{1}\) is connected directly across the generator, without the transformer. Show that \(R_{1}=\left(N_{\mathrm{p}} / N_{\mathrm{s}}\right)^{2} R_{2},\) by starting with \(\mathrm{Ohm}^{\prime}\) s law as applied to the secondary coil.
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Rotational Dynamics

Read Explanation

Electromagnetism

Read Explanation

Geometrical and Physical Optics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Classical Mechanics

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.