Please fully answer part A **Magnetic Force on a Charged Particle**A particle with a charge of \(-1.24 \times 10^{-8}\) C is moving with an instantaneous velocity given by:\[\vec{v} = (4.19 \times 10^4 \, \text{m/s}) \hat{i} + (-3.85 \times 10^4 \, \text{m/s}) \hat{j}\]**Part A***Problem Statement:*What is the force exerted on this particle by a magnetic field \(\vec{B} = (2.10 \, \text{T}) \hat{i}\)?*Task:*Express the \(x\), \(y\), and \(z\) components of the force in newtons, separated by commas.*Input Field:*\[ F_x, F_y, F_z = \boxed{} \, \text{N} \]This problem involves using the formula for the magnetic force on a moving charge: \(\vec{F} = q \, (\vec{v} \times \vec{B})\). Calculate the cross product of \(\vec{v}\) and \(\vec{B}\), then multiply by the charge \(q\) to find each component of the force.

When a charge particle q moving with velocity v→ enters in a space where magnetic field B→, a force F→ known as "Lorentz Force" acts on the charge particle and it is given by F → = q(v→ × B→) Fx i^ + Fy j^ + Fz k^ =q(v→ × B→) ...........................1 In equation 1, charge q is to be substituted with sign. Given Information:- charge q = -1.24×10-8 C velocity v→ = (4.19×104 m/s) i^ + (-3.85×104 m/s) j^ magnetic field B→ = (2.10 T) i^ We will now solve for the value of (v→ × B→) as shown below- v→ × B→ =i^j^k^4.19 × 104 m/s -3.85 × 104 m/s 02.10 T0 0 v→ × B→ = 0 i^ + 0j^ + 2.10×3.85 ×104 k^ T . m/s v→ × B→ = 0 i^ + 0 j^ + 8.085 × 104 k^ T . m/s .............................2 Putting the value of (v→ × B→) from equation 2 in equation 1 along with the value of charge q with sign, we get Fx i^ + Fy j^ +Fz k^ = -1.24 × 10-8 C (0 i^ + 0 j^ + 8.085 ×104 k^) T . m/s or Fx i^ + Fy j^ +Fz k^ = (0 i^ + 0 j^ - 10.0254 × 10-4 k^) C . T . m/s ⇒ Fx i^ + Fy j^ +Fz k^ = 0 i^ + 0 j^ - 10.0254 × 10-4 k^ Newton Hence the component of force are given by Fx = 0 NewtonFy = 0 NewtonFz = 10.0254 ×10-4 Newton along -k^ Thus Fx, Fy, Fz = (0,0,-10.0254×10-4) Newton is the final answer.