/*! 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} Q12E A horizontal rectangular surface... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A horizontal rectangular surface has dimensions 2.80cmby 3.20cmand is in a uniform magnetic field that is directed at an angle of 30.0°above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10×10-4Wb through the surface?

Short Answer

Expert verified

The magnitude of the magnetic field be to produce a flux of 3.10×10-4Wbthrough the surface is0.692T.

Step by step solution

01

Definition of magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

02

Determine the magnitude of the magnetic field

The magnetic flux can be calculated as

fB=∫B⇶Ä×dA→fB=∫BAcosθB=fBAcosθ

But

A=aba=2.30cmb=3.20cm

Than

B=fBabcosθ

Here a, b are the side of rectangle fBis the magnetic flux and B is the magnetic field.

Substitute all the value in the above equation.

B=fBabcosθB=3.10×10-4Wb2.80×10-2m3.20×10-2mcos60.0°B=0.692T

Hence, the magnitude of the magnetic field be to produce a flux of role="math" localid="1655717958537" 3.10×10-4Wbthrough the surface is 0.692T

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 91Ó°ÊÓ!

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 long, straight solenoid with a cross-sectional area of 8.00 cm2 is wound with 90 turns of wire per centimetre, and the windings carry a current of 0.350 A. A second winding of 12 turns encircles the solenoid at its centre. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400 s. What is the average induced emf in the second winding?

When switch Sin Fig. E25.29 is open, the voltmeter V reads 3.08 V. When the switch is closed, the voltmeter reading drops to 2.97 V, and the ammeter A reads 1.65 A. Find the emf, the internal resistance of the battery, and the circuit resistance R. Assume that the two meters are ideal, so they don’t affect the circuit.

Fig. E25.29.

In the circuit shown in Fig. E25.30, the 16.0-V battery is removed and reinserted with the opposite polarity, so that its negative terminal is now next to point a. Find (a) the current in the circuit (magnitude anddirection); (b) the terminal voltage Vbaof the 16.0-V battery; (c) the potential difference Vacof point awith respect to point c. (d) Graph the potential rises and drops in this circuit (see Fig. 25.20).

Two identical metal objects are mounted on insulating stands. Describe how you could place charges of opposite sign but exactly equal magnitude on the two objects.

In the circuit shown in Fig. E26.18,ε=36.V,R1=4.0Ω,R2=6.0Ω,R3=3.0Ω(a) What is the potential difference Vab between points a and b when the switch S is open and when S is closed? (b) For each resistor, calculate the current through the resistor with S open and with S closed. For each resistor, does the current increase or decrease when S is closed?

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.