# Chapter 3 - Vectors and Coordinates Systems - Exercises and Problems - Page 78: 21

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

We first find the magnitude of the vector; $B = \sqrt{B_x^2+B_y^2}$ $B = \sqrt{(2.0~T)^2+(-1.0~T)^2}$ $B = 2.2~T$ We then find the angle below the positive x-axis; $tan(\theta) = \frac{1.0}{2.0}$ $\theta = tan^{-1}(\frac{1.0}{2.0}) = 26.6^{\circ}$ 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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