Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 21, Problem 6P

A fixed 10.8-cm-diameter wire coil is perpendicular to a magnetic field 0.48 T pointing up. In 0.16 s,the field is changed to 0.25 T pointing down. What is the average induced emf in the coil?

Expert Solution & Answer
Check Mark
To determine

The average induced EMF.

Answer to Problem 6P

Solution:

The average induced EMF is equal to 41.75 mV.

Explanation of Solution

Given:

The magnetic field of the loop is 0.48 T.

The time taken to rotate the loop is 0.16 s.

The field is changed to 0.25 T.

The diameter of the loop is 10.8 cm.

Formula used:

The relation between the radius and the diameter is as follows:

r=d2

Here, d is the diameter.

The magnetic flux through the circular lop is as follows:

ϕ=BAcosθ

Here, B is the magnetic field, A is the area, and θ is the angle between the area and the magnetic field vector.

The expression for the EMF is as follows:

ε=dϕdt

Here, dϕ is the change in the flux and dt is the change in time.

Therefore, the equation ε=dϕdt is rewritten as follows:

ε=ϕ2ϕ1dt

Here, ϕ2 is the flux in the final position and ϕ1 is the flux in the initial situation.

Therefore,

ϕ1=BAcosθi

Here, θi is the initial angle.

And,

ϕ2=BAcosθf

Here, θf is the final angle.

Calculation:

Calculate the radius of the loop by substituting 10.8 cm for d in the equation r=d2.

r=10.8 cm2=10.8 cm(102 m1.00 cm)2=0.054 m

Now, substitute 0.054 m for r in the equation A=πr2.

A=π(0.054 m)2=(3.14)(0.054 m)2=9.16×103 m2

Now, calculate the magnetic flux in the initial condition by substituting 0.48 T for B, 9.16×103 m2 for A, and 0° for θi in the equation ϕ1=BAcosθi.

ϕ1=(0.48 T)(9.16×103 m2)cos0°=4.39×103 Wb

And, the magnetic flux in the final condition is obtained by substituting 0.25 T for B, 9.16×103 m2 for A, and 90° for θf in the equation ϕ2=BAcosθf.

ϕ2=(0.25 T)(9.16×103 m2)cos90°=2.29×103 Wb

Therefore, substitute 2.29×103 Wb for ϕ2, 4.39×103 Wb for ϕ1, and 0.16 s for dt in the equation ε=ϕ2ϕ1dt.

ε=4.39×103 Wb2.29×103 Wb0.16 s=41.75×103 V=41.75×103 V(1 mV1.00 V)=41.75 mV

Chapter 21 Solutions

Physics: Principles with Applications

Ch. 21 - Prob. 11QCh. 21 - Prob. 12QCh. 21 - Prob. 13QCh. 21 - Prob. 14QCh. 21 - Prob. 15QCh. 21 - Prob. 16QCh. 21 - Prob. 17QCh. 21 - Prob. 18QCh. 21 - Prob. 19QCh. 21 - Prob. 20QCh. 21 - Prob. 21QCh. 21 - Prob. 22QCh. 21 - Prob. 23QCh. 21 - Prob. 24QCh. 21 - Prob. 1PCh. 21 - Prob. 2PCh. 21 - The rectangular loop in Fig. 21-5813 is being...Ch. 21 - If the solenoid in Fig. 21-59 |D is being pulled...Ch. 21 - An 18.5-cm-diameter loop of wire is initially...Ch. 21 - A fixed 10.8-cm-diameter wire coil is...Ch. 21 - A 16-cm-diameter circular loop of wire is placed...Ch. 21 - Prob. 8PCh. 21 - Prob. 9PCh. 21 - A circular loop in the plane of the paper lies in...Ch. 21 - Prob. 11PCh. 21 - Prob. 12PCh. 21 - Prob. 13PCh. 21 - Prob. 14PCh. 21 - Prob. 15PCh. 21 - Prob. 16PCh. 21 - Prob. 17PCh. 21 - Prob. 18PCh. 21 - Prob. 19PCh. 21 - Prob. 20PCh. 21 - Prob. 21PCh. 21 - A simple generator has a square armature 6.0 cm on...Ch. 21 - Prob. 23PCh. 21 - Prob. 24PCh. 21 - Prob. 25PCh. 21 - Prob. 26PCh. 21 - Prob. 27PCh. 21 - Prob. 28PCh. 21 - Prob. 29PCh. 21 - Prob. 30PCh. 21 - Prob. 31PCh. 21 - Prob. 32PCh. 21 - Prob. 33PCh. 21 - Prob. 34PCh. 21 - Prob. 35PCh. 21 - Prob. 36PCh. 21 - Prob. 37PCh. 21 - Prob. 38PCh. 21 - Prob. 39PCh. 21 - Prob. 40PCh. 21 - Prob. 41PCh. 21 - Prob. 42PCh. 21 - Prob. 43PCh. 21 - Prob. 44PCh. 21 - Prob. 45PCh. 21 - Prob. 46PCh. 21 - Prob. 47PCh. 21 - Prob. 48PCh. 21 - Prob. 49PCh. 21 - Prob. 50PCh. 21 - Prob. 51PCh. 21 - Prob. 52PCh. 21 - Prob. 53PCh. 21 - Prob. 54PCh. 21 - Prob. 55PCh. 21 - Prob. 56PCh. 21 - (a) What is the reactance of a well-insulated...Ch. 21 - Prob. 58PCh. 21 - Prob. 59PCh. 21 - Prob. 60PCh. 21 - Prob. 61PCh. 21 - Determine the total impedance, phase angle, and...Ch. 21 - Prob. 63PCh. 21 - Prob. 64PCh. 21 - Prob. 65PCh. 21 - Prob. 66PCh. 21 - Prob. 67PCh. 21 - The variable capacitor in the tuner of an AM radio...Ch. 21 - Prob. 69PCh. 21 - A resonant circuit using a 260-nF capacitor is to...Ch. 21 - Prob. 71PCh. 21 - Prob. 72GPCh. 21 - Prob. 73GPCh. 21 - Prob. 74GPCh. 21 - Prob. 75GPCh. 21 - Prob. 76GPCh. 21 - Power is generated at 24 kV at a generating plant...Ch. 21 - Prob. 78GPCh. 21 - Prob. 79GPCh. 21 - Prob. 80GPCh. 21 - Prob. 81GPCh. 21 - Prob. 82GPCh. 21 - Prob. 83GPCh. 21 - Prob. 84GPCh. 21 - Prob. 85GPCh. 21 - Prob. 86GPCh. 21 - Prob. 87GPCh. 21 - Prob. 88GPCh. 21 - Prob. 89GP
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