Given G 20a r 50a 40a , at (1, 2, 6) the component of G perpendicular to surface 2 is (a) 20a r (b) 50a (c) 40a (d) 20a r 40a (e) 40a 20a ...

The correct answer is e 40a 20a This is because at 1 2 6 the component o ...

Given G = 20a r + 50a + 40a , at (1, /2, /6) the

Question:

Given G = 20a_r+ 50a_θ + 40a_φ, at (1, π/2, π/6) the component of G perpendicular to surface θ = π/2 is
(a) 20a_r
(b) 50a_θ
(c) 40a_φ
(d) 20a_r + 40a_θ
(e) - 40a_θ + 20a_φ

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Elements Of Electromagnetics

ISBN: 9780190698614

7th Edition

Authors: Matthew Sadiku

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Question Posted: Jan 09, 2023 07:10 AM

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Given G = 20a r + 50a + 40a , at (1, /2, /6) the

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The correct answer is e 40a 20a This is because at 1 2 6 the component o...View the full answer

Elements Of Electromagnetics

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