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

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