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 nnevermind its 3.8 n

3.8 N

Explanation:

The force, F exerted on a charge q moving with a velocity, v in a magnetic field of strength, B is F = Bqv

Given that B = 3.0 × 10² T , q = 2.5 μC = 2.5 × 10⁻⁶ C and v = 5.0 × 10³ m/s.

So, F = Bqv = 3.0 × 10² T × 2.5 × 10⁻⁶ C × 5.0 × 10³ m/s = 37.5 × 10⁻¹ N = 3.75 N

F = 3.75 N ≅ 3.8 N

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

there are no contact forces mentioned on the list of choices that you've provided.