Answer:
The amount of current that must flow through the wire for it to be suspended against gravity by magnetic force = 6.125 A
Explanation:
Force on a wire carrying current in an electric field is given by
F = (B)(I)(L) sin θ
For this question,
The magnetic force must match the weight of the wire.
F = mg
mg = (B)(I)(L) sin θ
(m/L)g = (B)(I) sin θ
Mass per unit length = 75 g/m = 0.075 kg/m
B = magnetic field = 0.12 T
I = ?
g = acceleration due to gravity = 9.8 m/s
θ = angle between wire's current direction and magnetic field = 90°
0.075 × 9.8 = 0.12 × I sin 90°
I = 0.075 × 9.8/0.12 = 6.125 A
To determine the displacement, since we are given the potential energy, we use the equation for potential energy. For a spring, it is one-half the product of the spring constant and the square of the displacement. We do as follows:
PE = kx^2/2
5 Nm = 50N/m (x^2)
x = 0.32 m
Therefore, the displacement would be 0.32 m.
The instant it was dropped, the ball had zero speed.
After falling for 1 second, its speed was 9.8 m/s straight down (gravity).
Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.
Falling for 1 second at an average speed of 4.9 m/s, is covered <em>4.9 meters</em>.
ANYTHING you drop does that, if air resistance doesn't hold it back.
