A particle with a charge of 5.13 μC has a velocity of 8.64 x106 m/s in a direction perpendicular to a magnetic fieldof 1.99 x 10-4 T. The magnitudeof the force on this particle is 1 Newtons?

F = 0.009 N

Explanation:

Given that

Charge, q = 5.13 μC

Velocity, V = 8.64 x 10⁶ m/s

Magnetic field, B = 1.99 x 10⁻⁴ T

The force on a charge q moving with velocity v is given as follows

F = q V B

Now by putting the values in the above equation we get

[tex]F = 5.13/times 10^{-6}/times 8.64/times 10^{6}/times 1.99/times 10^{-4} / N [/tex]

F=0.00882 N

F = 0.009 N

Therefore the force on the particle will be 0.009 N.