Answer:
Explanation: To solve this problem, we can use the equation for the force on a current-carrying conductor in a magnetic field:
F = I L x B
where F is the force on the conductor, I is the current flowing through the conductor, L is the length of the conductor, and B is the magnetic field strength.
a) Just after the switch is closed, the circuit is completed and a current starts to flow in the bar. The emf of the battery causes a current to flow in the circuit, which is given by Ohm's Law:
I = V / R = 12 / 5 = 2.4 A
The direction of the current is from the battery, through the bar, and back to the battery. Since the magnetic field is directed into the plane, the force on the bar is perpendicular to both the magnetic field and the current. Therefore, the bar experiences a sideways force that pushes it along the rails. The magnitude of the force is given by:
F = I L B = 2.4 x 0.36 x 2.4 = 2.0736 N
The mass of the bar is 0.90 kg, so the acceleration of the bar is:
a = F / m = 2.0736 / 0.90 = 2.304 m/s^2
Therefore, the acceleration of the bar just after the switch is closed is 2.304 m/s^2.
b) When the bar's speed is 2.0 m/s, the force on the bar is still given by:
F = I L B = 2.4 x 0.36 x 2.4 = 2.0736 N
However, now there is an additional force acting on the bar due to its motion through the air. This force is given by:
F_air = -0.5 ρ C_d A v^2
where ρ is the density of air, C_d is the drag coefficient, A is the cross-sectional area of the bar, and v is the speed of the bar. We can estimate the cross-sectional area of the bar as A = 0.01 m^2, and assume that the drag coefficient is C_d = 1. The density of air is ρ = 1.2 kg/m^3.
Substituting the given values, we get:
F_air = -0.5 x 1.2 x 1 x 0.01 x 2^2 = -0.024 N
The negative sign indicates that the air resistance is acting in the opposite direction to the motion of the bar.
The net force on the bar is therefore:
F_net = F + F_air = 2.0736 - 0.024 = 2.0496 N
Using Newton's Second Law, we can calculate the acceleration of the bar:
a = F_net / m = 2.0496 / 0.90 = 2.2778 m/s^2
Therefore, the acceleration of the bar when its speed is 2.0 m/s is 2.2778 m/s^2.
c) The bar's terminal speed is reached when the air resistance force is equal in magnitude and opposite in direction to the force due to the magnetic field. At this point, the net force on the bar is zero, so the acceleration is zero and the bar moves at a constant speed.
Setting F_net = 0, we can solve for the terminal speed:
F + F_air = 0
I L B - 0.5 ρ C_d A v^2 = 0
2.4 x 0.36 x 2.4 - 0.5
ρ C_d A v^2 = 2.0736
v^2 = (2.0736) / (ρ C_d A)
v^2 = (2.0736) / (1.2 x 1 x 0.01)
v^2 = 172.8
v = sqrt(172.8)
v = 13.142 m/s
Therefore, the bar's terminal speed is 13.142 m/s.