Finna Learn! **Problem Statement:**A particle with a charge of \( -1.24 \times 10^{-8} \, \text{C} \) is moving with instantaneous velocity \[\vec{v} = (4.19 \times 10^{4} \, \text{m/s})\hat{i} + (-3.85 \times 10^{4} \, \text{m/s})\hat{j}.\]**Part A:**What is the force exerted on this particle by a magnetic field \[\vec{B} = (2.10 \, \text{T}) \hat{i}?\]**Instructions:**Express the \( x, y, \) and \( z \) components of the force in newtons separated by commas.**Input Field:**\( F_x, F_y, F_z = \) [Input Box] N**Submit Button:** Options to submit your answer with the ability to view previous answers or request further assistance. **Part B**What is the force exerted on this particle by a magnetic field \(\vec{B} = (2.10 \, \text{T}) \, \hat{k}\)?Express the \(x\), \(y\), and \(z\) components of the force in newtons separated by commas.\[ F_x, F_y, F_z = \, \_\_\_\_\_\_ \, \text{N} \]Buttons are available for mathematical expressions and formatting options. The input section is where you enter the force components. After inputting, you can click "Submit" to check your answer or "Request Answer" for guidance.

The formula force on a charge q moving in a magnetic field B→ with a velocity v→ is F→=q(v→×B→) Using this formula we will solve the given problem.Given: q=-1.24×10-8 Cv→=4.9×104 m/si^+(-3.85×104 m/s)j^B→=(2.10 T)i^ Therefore, the force on the charge is: F→=q(v→×B→)F→=(-1.24×10-8 C)[4.9×104 m/si^+(-3.85×104 m/s)j^]×(2.10 T)i^F→=0i^+1.0025×10-3(-k^)F→=0i^+0j^+(-1.0025×10-3N k^) Hence, Fx=0Fy=0Fz=-0.0010025 N

Science

Physics

A particle with a charge of –1.24x10 8 Cis moving with instantaneous velocity v = (4.19x10 m/s)i + (-3.85x104 m/s)j Part A What is the force exerted on this particle by a magnetic field B = (2.10 T) i? Express the a, y, and z components of the force in newtons separated by commas Πνα ΑΣφ F,, F,. F. = N