A particle with unit charge \( (q=1) \) enters a constant magnetic field \( \mathbf{B}=\mathbf{i}+j \) with velocity \( \mathbf{v}=20 \mathbf{k} \). Find the magnitude and direction of the force on the particle. Make a sketch of the magnetic field, the velocity, and the force. What is the magnitude of the force on the particle? (Type an exact answer, using radicals as needed.)

To find the force on a charged particle moving in a magnetic field, we can use the Lorentz force equation: \[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \] Given that the charge \( q = 1 \), the velocity \( \mathbf{v} = 20 \mathbf{k} \), and the magnetic field \( \mathbf{B} = \mathbf{i} + \mathbf{j} \), we can calculate the cross product \( \mathbf{v} \times \mathbf{B} \). First, we can express the vectors in component form: \[ \mathbf{v} = 0\mathbf{i} + 0\mathbf{j} + 20\mathbf{k} \] \[ \mathbf{B} = 1\mathbf{i} + 1\mathbf{j} + 0\mathbf{k} \] Now, we calculate the cross product: \[ \mathbf{v} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 0 & 20 \\ 1 & 1 & 0 \end{vmatrix} \] Calculating the determinant, we find: \[ \mathbf{v} \times \mathbf{B} = \mathbf{i} (0 \cdot 0 - 20 \cdot 1) - \mathbf{j} (0 \cdot 0 - 20 \cdot 1) + \mathbf{k} (0 \cdot 1 - 0 \cdot 0) \] This simplifies to: \[ \mathbf{v} \times \mathbf{B} = -20 \mathbf{i} + 20 \mathbf{j} + 0\mathbf{k} = -20 \mathbf{i} + 20 \mathbf{j} \] Thus, the force on the particle is: \[ \mathbf{F} = \mathbf{v} \times \mathbf{B} = -20 \mathbf{i} + 20 \mathbf{j} \] To find the magnitude of the force, we can calculate: \[ |\mathbf{F}| = \sqrt{(-20)^2 + (20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \] So, the magnitude of the force on the particle is \( 20\sqrt{2} \). In terms of direction, the force vector \( \mathbf{F} = -20 \mathbf{i} + 20 \mathbf{j} \) indicates that the force is directed at an angle of \( \tan^{-1}(-1) \) with respect to the negative \( x \)-axis, which points up towards the \( y \)-direction (the second quadrant). To summarize, the magnitude of the force on the particle is \( 20\sqrt{2} \) and its direction is in the quadrant between negative x and positive y.