University Physics Volume 2
University Physics Volume 2
18th Edition
ISBN: 9781938168161
Author: OpenStax
Publisher: OpenStax
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Chapter 13, Problem 13.1CYU

Chapter 13, Problem 13.1CYU, Chek sour Understanding A closely und coil has a radius of 4.0 cm. 50 turns, and a total ieslstancc Chek sour Understanding A closely und coil has a radius of 4.0 cm. 50 turns, and a total ieslstancc of 40 . At what rae must a magnetic field peipendicular lathe face of the coil change En euler to prodice Joule hearing in the cull a a tale of 2.0 mW

Expert Solution & Answer
Check Mark
To determine

To Calculate:

The rate at which magnetic field perpendicular to the face of the coil carrying current changes in order to produce the given Joule heating in the coil.

Answer to Problem 13.1CYU

The rate change of the magnetic field is 1.1 T/s.

Explanation of Solution

Given:

Radius of the coil, r=4.0cm = 0.04 m

Number of turns, N=50

Total resistance, R=40Ω

Joule heating in the coil that is power, P=2mW=2×103W

Formula used:

Power loss in a circuit carrying current is, P=I2R ----(1).

Here R is the resistance of the wire.

Emf produced in a coil is equal to negative of rate change of flux, ε=dφdt ----(2).

Here φ is magnetic flux which is given as φ=nB.A ----(3)

Where, n is the number of turns, B is the magnetic field & A is the area of the coil.

From equation(1) current in the coil is given as,

  I=PR

Substituting the value of P & R we get,

  I=PR=2× 10 340=7.0×103=7.0mA

And also power P=VI or the emf is, emf=V=PI=2×1037×103=0.28V

From equation (2) and equation (3), the emf produced in the coil is ε=AdBdt. Here A is the area of the coil which is A=πr2. Now

  ε=nπr2dBdt

So, the rate at which magnetic field is changing is

  dBdt=εnπr2

Substituting the value of ε, π & r, we get,

  dBdt=0.2850×(3.14)(0.04)2=1.1 T/s

Conclusion:

Thus, the rate at which the magnetic field perpendicular to the face of the coil carrying current should change is 1.1 T/s.

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Chapter 13 Solutions

University Physics Volume 2

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