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

battery with hi vtge high val resist in series

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.