/*! 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} Q. 11 I聽I The metal equilateral trian... [FREE SOLUTION] | 91影视

91影视

I

I The metal equilateral triangle in FIGURE EX30.11,20cmon each side, is halfway into a0.10Tmagnetic field.

a. What is the magnetic flux through the triangle?

b. If the magnetic field strength decreases, what is the direction of the induced current in the triangle?

Short Answer

Expert verified

(a) The magnetic flux is(a)m=8.6104Wb.

(b) The induced current flows clockwise.

Step by step solution

01

Step1:Definition of magnetic field

There is a magnetic field. Magnetic fields are vector fields that describe the magnetic influence of currents and magnetised materials. Magnetic fields are frequently seen in everyday life in the form of permanent magnets, which attract or repel other magnets by pulling on magnetic materials (such as iron).

02

step2:Find height(part a)

(a) Magnetic flux is the amount of magnetic field that flows through a loop in area A. Is when magnet is parallel to the normal of a plane, the magnetic flux is given

m=BA

The triangle's area is equal to half of its base multiplied by its height. The height is derived from

h=(20cm)sin60=17.3cm

So, we get the area of the triangle by

A=12bh=12(0.20m)17.3102m=1.73102m2

03

step3:Find area of the flux

Because the magnetic field flows through half of the total loop, the flux area is

A=121.73102m2=8.65103m2

Now, we plug the values forlocalid="1648879727201" A,=0and B into equation (2) to get m

m=BAcos=(0.10T)8.65103m2cos0

04

Find direction(part b)

(b) According to Lenz's law, the direction of the induced emf drives current around a wire loop in the opposite direction of the change in magnetic flux that causes the emf. When the applied magnetic field's strength is reduced, the induced magnetic field points in the same direction as the applied magnetic field B. As a result, the induced magnetic field is pointing into the page.

To determine the direction of the induced current, apply the right-hand rule. Your four wrapped fingers represent the direction of the induced magnetic field (into the page), and your thumb represents the direction of the current.

(a)m=8.6104Wb.

(b) The induced current flows clockwise.

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!

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

The resistance of the loop in FIGURE EX30.15 is 0.20Is the magnetic field strength increasing or decreasing? At what rate T/s ?

10. An inductor with a2.0A current stores energy. At what current will the stored energy be twice as large?

A 3.0-cm-diameter, 10-turn coil of wire, located at z=0in thexy-plane, carries a current of 2.5A.A2.0-mm-diameter conducting loop with 2.0*10-4resistance is also in the xyplane at the center of the coil. At t=0s, the loop begins to move along the z-axis with a constant speed of 75m/s. What is the induced current in the conducting loop att=200ms?The diameter of the conducting loop is much smaller than that of the coil, so you can assume that the magnetic field through the loop is everywhere the on-axis field of the coil.

A 40-turn, 4.0-cm-diameter coil with R=0.40 surrounds a 3.0-cm-diameter solenoid. The solenoid is 20 cm long and has 200 turns. The

60 Hzcurrent through the solenoid is I = I0sin(2ft). What is I0 if the maximum induced current in the coil is 0.20A??

Your camping buddy has an idea for a light to go inside your tent. He happens to have a powerful and heavy horseshoe magnet that he bought at a surplus store. This magnet creates a 0.20Tfield between two pole tips 10cmapart. His idea is to build the hand-cranked generator shown in FIGURE .He thinks you can make enough current to fully light a 1.0lightbulb rated at 4.0W. That鈥檚 not super bright, but it should be plenty of light for routine activities in the tent.

a. Find an expression for the induced current as a function of time if you turn the crank at frequency f. Assume that the semicircle is at its highest point at t=0s.

b. With what frequency will you have to turn the crank for the maximum current to fully light the bulb? Is this feasible?

See all solutions

Recommended explanations on Physics Textbooks

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.