Question

A charge q = 40 μC moves with an instantaneous velocity v = 5×10 m/s through the uniform fields E = 6×10(0.52i + 0.56j + 0.645k) V/m and B = 1.1781i +1.02j + 0.68k T. Find the magnitude and direction of the instantaneous force on q.

Answer 1

The force on a charge moving through electric and magnetic fields is determined using Coulomb's law and the Lorentz force equation. Vector cross products calculate the force, and the right-hand rule confirms the direction for the magnetic component.

Step-by-step explanation:

The student asks about the magnitude and direction of the instantaneous force on a charge moving through electric and magnetic fields. To find the electric force (F), we use Coulomb's law: F = qE, where q is the charge and E is the electric field. For a magnetic force, we utilize the Lorentz force equation: F = q(v x B), where v is the charge's velocity and B is the magnetic field. Calculations are performed using vector cross products, and the direction of the magnetic force is determined using the right-hand rule.

The electric force is given by:

• F = qE = (40× 10µC)(6× 10(0.52i + 0.56j + 0.645k) V/m)

The magnetic force can be found by taking the cross product of v and B vectors:

• F = q(v x B)

The final step involves summing the electric and magnetic forces to find the net force on the charge.