#### 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.