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
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.
