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

nevermind its 3.8 N

Answer 1

3.8 is the correct answer!

Answer 2

3.8 N

Step-by-step 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