Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 20, Problem 1P
To determine

Part (a) To Determine:

Force per metre length of straight wire carrying current 6.40 A placed perpendicular to Uniform magnetic field of 0.90 T.

Expert Solution
Check Mark

Answer to Problem 1P

Solution:

Force per metre length of straight wire = 5.76 N/m

Explanation of Solution

Given:

Current I = 6.40 A; Magnetic field B = 0.90 T and Ϙ = 90o

Formula Used:

F=IlBSinθ

Calculation:

In order to find the value of force per metre length of current carrying conductor, we have to use the equation,

F=IlBSinθ

For force per unit length equation becomes,

Force per unit length = IBSinθ

Putting the values in this equation we get,

Force per unit length = 6.40×0.90×Sin90

= 6.40×0.90×1

= 5.76N/m

To determine

Part (b) To Determine:

Force per metre length of straight wire carrying current 6.40A placed 350 to Uniform magnetic field of 0.90T

Expert Solution
Check Mark

Answer to Problem 1P

Solution:

Force per metre length of straight wire = 3.28 N/m

Explanation of Solution

Given:

Current I=6.40A; Magnetic field B= 0.90 T and Ϙ=350

Formula Used:

F=ILBSinθ

Calculation:

To find the value of force per metre length of current carrying conductor, we have to use the equation,

F=ILBSinθ

For force per unit length equation becomes

Force per unit length becomes = IBSinθ

=(6.40)(0.90)Sin350

=(6.40)(0.90)(0.57)

=3.28N/m

Chapter 20 Solutions

Physics: Principles with Applications

Ch. 20 - Prob. 11QCh. 20 - Prob. 12QCh. 20 - 13. Explain why a strong magnet held near a CRT...Ch. 20 - Prob. 14QCh. 20 - Prob. 15QCh. 20 - Prob. 16QCh. 20 - Prob. 17QCh. 20 - If a moving charged particle is deflected sideways...Ch. 20 - Prob. 19QCh. 20 - Prob. 20QCh. 20 - Prob. 21QCh. 20 - Prob. 22QCh. 20 - Prob. 23QCh. 20 - Why will either pole of a magnet attract an...Ch. 20 - Prob. 25QCh. 20 - Prob. 26QCh. 20 - Prob. 27QCh. 20 - Prob. 28QCh. 20 - Prob. 29QCh. 20 - Prob. 30QCh. 20 - Prob. 31QCh. 20 - Prob. 1PCh. 20 - Prob. 2PCh. 20 - A 240-m length of wire stretches between two...Ch. 20 - Prob. 4PCh. 20 - Prob. 5PCh. 20 - Prob. 6PCh. 20 - Prob. 7PCh. 20 - Prob. 8PCh. 20 - Prob. 9PCh. 20 - Prob. 10PCh. 20 - Prob. 11PCh. 20 - Find the direction of the force on a negative...Ch. 20 - Prob. 13PCh. 20 - Prob. 14PCh. 20 - Prob. 15PCh. 20 - Prob. 16PCh. 20 - Prob. 17PCh. 20 - Prob. 18PCh. 20 - Prob. 19PCh. 20 - Prob. 20PCh. 20 - Prob. 21PCh. 20 - Prob. 22PCh. 20 - Prob. 23PCh. 20 - Prob. 24PCh. 20 - Prob. 25PCh. 20 - Prob. 26PCh. 20 - Prob. 27PCh. 20 - Prob. 28PCh. 20 - Prob. 29PCh. 20 - Prob. 30PCh. 20 - Prob. 31PCh. 20 - Prob. 32PCh. 20 - Prob. 33PCh. 20 - Prob. 34PCh. 20 - Prob. 35PCh. 20 - Prob. 36PCh. 20 - Prob. 37PCh. 20 - Prob. 38PCh. 20 - Prob. 39PCh. 20 - Prob. 40PCh. 20 - Prob. 41PCh. 20 - Prob. 42PCh. 20 - Prob. 43PCh. 20 - Prob. 44PCh. 20 - Prob. 45PCh. 20 - Prob. 46PCh. 20 - Prob. 47PCh. 20 - Prob. 48PCh. 20 - Prob. 49PCh. 20 - Prob. 50PCh. 20 - Prob. 51PCh. 20 - Prob. 52PCh. 20 - Prob. 53PCh. 20 - A circular coil 12.0 cm in diameter and containing...Ch. 20 - Prob. 55PCh. 20 - Prob. 56PCh. 20 - Prob. 57PCh. 20 - Prob. 58PCh. 20 - Prob. 59PCh. 20 - Prob. 60PCh. 20 - Prob. 61PCh. 20 - Prob. 62PCh. 20 - Prob. 63PCh. 20 - Prob. 64PCh. 20 - Prob. 65PCh. 20 - Prob. 66PCh. 20 - Prob. 67GPCh. 20 - Prob. 68GPCh. 20 - Prob. 69GPCh. 20 - Prob. 70GPCh. 20 - Prob. 71GPCh. 20 - Prob. 72GPCh. 20 - Prob. 73GPCh. 20 - Prob. 74GPCh. 20 - Prob. 75GPCh. 20 - Prob. 76GPCh. 20 - Prob. 77GPCh. 20 - Prob. 78GPCh. 20 - Prob. 79GPCh. 20 - Prob. 80GPCh. 20 - Prob. 81GPCh. 20 - Prob. 82GPCh. 20 - Prob. 83GPCh. 20 - Prob. 84GPCh. 20 - Prob. 85GPCh. 20 - Prob. 86GPCh. 20 - Prob. 87GP
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