Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 22, Problem 42GP
To determine

The values of E0 and B0 .

Expert Solution & Answer
Check Mark

Answer to Problem 42GP

The answers are 37.7V/m and 1.26×107T .

Explanation of Solution

Given data:

Power, P0=95W

Radius, r=2m

Speed of the light, c=3×108m/s

Permittivity of vacuum, ε0=8.85×1012C2/N.m2

Formula used:

It is known that: E02=2P0(4πr2)(ε0c) and B0=E0c

Where,

  P is the power.

  c is speed of the light.

  E is the electric field.

  ε0 is the permittivity of vacuum.

  r is the radius.

  B0 is the magnetic field.

Calculation:

Substitute the given data into the formula E02=2P0(4πr2)(ε0c) to find a electric field.

  E0=2P0(4πr2)(ε0c)=2×95W4π(2.00m)2(8.85×1012C2/N.m2×3×108m/s)=37.7V/m

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