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

Given:

B = $3 \times 10^{2}$ T

V=$5 \times 10^{3} \frac{m}{s}$

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 = $2.5 \times 10^{-6} \times 3 \times 10^{2} \times 5 \times 10^{3}$

F = 37.5 × $10^{-1}$

F = 3.75 Newton

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

Answer 2

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