Question

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?

Answer 1

Answer:

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.