Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 3 - Vectors and Coordinate Systems - Exercises and Problems: 9

Answer

The magnitude of the vector is 2.2 T and the direction is an angle of $26.6^{\circ}$ below the positive x-axis.

Work Step by Step

(a) We can find the magnitude of the vector as: $B = \sqrt{B_x^2+B_y^2}$ $B = \sqrt{(2.0~T)^2+(-1.0~T)^2}$ $B = 2.2~T$ We can find the angle below the positive x-axis as: $tan(\theta) = \frac{1.0}{2.0}$ $\theta = tan^{-1}(\frac{1.0}{2.0}) = 26.6^{\circ}$ The magnitude of the vector is 2.2 T and the direction is an angle of $26.6^{\circ}$ below the positive x-axis.
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