A particle with unit charge ( \( q=1 \) ) enters a constant magnetic field \( \mathbf{B}=\mathbf{i}+j \) with velocity \( \mathbf{v}=20 \mathrm{k} \). 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? \( 20 \sqrt{2} \) (Type an exact answer, using radicals as needed.) What is the direction of the force on the particle? Select the correct choice below and fill in the answer box to complete your choice. A. The force is applied at \( a(n) \square^{\circ} \) angle with the positive \( y \)-axis in the \( y z \)-plane. B. The force is applied at \( a(n) \square^{\circ} \) angle with the positive \( x \)-axis in the xz-plane. C. The force is applied at \( a(n) \square^{\circ} \) angle with the positive \( x \)-axis in the \( x y \)-plane.

To find the force on a charged particle moving in a magnetic field, we can use the equation for the magnetic force: \[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \] Given \( q = 1 \), \( \mathbf{B} = \mathbf{i} + \mathbf{j} \), and \( \mathbf{v} = 20 \mathbf{k} \), we can compute the cross product \( \mathbf{v} \times \mathbf{B} \). Calculating the cross product, we have: \[ \mathbf{v} \times \mathbf{B} = (20 \mathbf{k}) \times (\mathbf{i} + \mathbf{j}) = 20 (\mathbf{k} \times \mathbf{i}) + 20 (\mathbf{k} \times \mathbf{j}) \] Using the right-hand rule, we find \( \mathbf{k} \times \mathbf{i} = -\mathbf{j} \) and \( \mathbf{k} \times \mathbf{j} = \mathbf{i} \). Thus: \[ \mathbf{v} \times \mathbf{B} = 20 (-\mathbf{j}) + 20 \mathbf{i} = 20 \mathbf{i} - 20 \mathbf{j} \] This means the force \( \mathbf{F} \) is: \[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) = 20 \mathbf{i} - 20 \mathbf{j} \] The magnitude of this force is: \[ |\mathbf{F}| = \sqrt{(20)^2 + (-20)^2} = \sqrt{400 + 400} = 20 \sqrt{2} \] Now, regarding the direction of the force: To analyze the angle, we can use the components of the force \( \mathbf{F} = (20, -20, 0) \). We can find the angle with respect to the axis in the \( xy \)-plane. The \( x \)-component is 20 and the \( y \)-component is -20. The angle \( \theta \) with the positive \( x \)-axis is given by: \[ \tan(\theta) = \frac{F_y}{F_x} = \frac{-20}{20} = -1 \] Thus, \( \theta = -45^\circ \), indicating it points downward in the \( xy \)-plane. The angle is typically expressed positively. Therefore, the answer to the force angle is \( 315^\circ \) (or \( -45^\circ \)), which is measured from the positive x-axis. So for the final answers: - The magnitude of the force on the particle is \( 20\sqrt{2} \). - The force is applied at \( 315^{\circ} \) angle with the positive \( x \)-axis in the \( xy \)-plane (choose option C).