Question

What is the force exerted on a charge of 2.5 µC moving perpendicular through a magnetic field of 3.0 × 102 T with a velocity of 5.0 × 103 m/s?

3.8 N
38 N
3.8 × 105 N
3.8 × 106 N

Answer 1

It's 3.8 N - just took it online and got it right. 

Answer 2

Given:

B =
`3 * 10^(2)` T

V=
`5 * 10^(3) (m)/(s)`

q = 2.5 ×
`10^(-6)` C

α = 90

To find:

Force = ?

Formula used:

Force on the moving charge is given by,

F = q V B sin α

Where F = force exerted on moving charge

V = velocity of charge

q = charge

α = angle between direction of V and B

Solution:

F = q V B sin α

Where F = force exerted on moving charge

V = velocity of charge

q = charge

α = angle between direction of V and B

F =
`2.5 * 10^(-6) * 3 * 10^(2) * 5 * 10^(3)`

F = 37.5 ×
`10^(-1)`

F = 3.75 Newton

Thus, the force acting on the moving charge is 3.75 Newton.