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 Lorentz force equation: \[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \] where: - \( \mathbf{F} \) is the force, - \( q \) is the charge of the particle, - \( \mathbf{v} \) is the velocity vector, - \( \mathbf{B} \) is the magnetic field vector. ### Step 1: Identify the vectors Given: - Charge \( q = 1 \) - Magnetic field \( \mathbf{B} = \mathbf{i} + \mathbf{j} \) - Velocity \( \mathbf{v} = 20 \mathbf{k} \) ### Step 2: Calculate the cross product \( \mathbf{v} \times \mathbf{B} \) We can express the vectors in component form: - \( \mathbf{v} = (0, 0, 20) \) - \( \mathbf{B} = (1, 1, 0) \) 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: \[ \mathbf{v} \times \mathbf{B} = \mathbf{i} \begin{vmatrix} 0 & 20 \\ 1 & 0 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 0 & 20 \\ 1 & 0 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 0 & 0 \\ 1 & 1 \end{vmatrix} \] Calculating each of the 2x2 determinants: 1. \( \begin{vmatrix} 0 & 20 \\ 1 & 0 \end{vmatrix} = (0)(0) - (20)(1) = -20 \) 2. \( \begin{vmatrix} 0 & 20 \\ 1 & 0 \end{vmatrix} = -20 \) 3. \( \begin{vmatrix} 0 & 0 \\ 1 & 1 \end{vmatrix} = (0)(1) - (0)(1) = 0 \) Putting it all together: \[ \mathbf{v} \times \mathbf{B} = -20 \mathbf{i} + 20 \mathbf{j} + 0 \mathbf{k} = -20 \mathbf{i} + 20 \mathbf{j} \] ### Step 3: Calculate the force \( \mathbf{F} \) Now, substituting back into the Lorentz force equation: \[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) = 1 \cdot (-20 \mathbf{i} + 20 \mathbf{j}) = -20 \mathbf{i} + 20 \mathbf{j} \] ### Step 4: Find the magnitude of the force The magnitude of the force is given by: \[ |\mathbf{F}| = \sqrt{(-20)^2 + (20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \] ### Step 5: Determine the direction of the force The force vector \( \mathbf{F} = -20 \mathbf{i} + 20 \mathbf{j} \) indicates that the force has components in the negative \( x \)-direction and positive \( y \)-direction. To find the angle with respect to the positive \( y \)-axis in the \( y z \)-plane, we can use the tangent function: \[ \tan(\theta) = \frac{|\text{component in } x|}{|\text{component in } y|} = \frac{20}{20} = 1 \] Thus, \( \theta = 45^\circ \). ### Final Answers 1. **Magnitude of the force**: \( 20\sqrt{2} \) 2. **Direction of the force**: - A. The force is applied at \( 45^\circ \) angle with the positive \( y \)-axis in the \( y z \)-plane. ### Sketch - The magnetic field \( \mathbf{B} \) can be represented as a vector in the \( x-y \) plane. - The velocity \( \mathbf{v} \) is along the \( z \)-axis. - The force \( \mathbf{F} \) points in the negative \( x \)-direction and positive \( y \)-direction, forming a vector in the \( x-y \) plane.