Question

A bar of length L = 0.36m is free to slide without friction on horizontal rails. A uniform magnetic field B = 2.4T is directed into the plane. At one end of the rails there is a battery with emf = 12V and a Switch S. The bar has the mass 0.90kg and resistance 5.0ohm. ignore all the other resistance in the circuit. The switch is closed at time t = 0. a) Just after the switch is closed, what is the acceleration of the bar? b)what is the acceleration of. the bar when its speed is 2.0m/s? c) what is the bar's terminal speed?

Answer 1

The acceleration of the bar just after the switch is closed is 13.33 m/s^2. The acceleration of the bar when its speed is 2.0 m/s is 6.00 m/s^2. The bar's terminal speed is 13.33 m/s.

Step-by-step explanation:

(a) The acceleration of the bar just after the switch is closed can be determined using the equation:

a = (E - Bvd)/m

where E is the emf of the battery, B is the magnetic field, v is the velocity of the bar, d is the length of the bar, and m is the mass of the bar. Plugging in the given values, we get:

a = (12 - 2.4 * 0 * 0.36) / 0.90 = 13.33 m/s^2

(b) The acceleration of the bar when its speed is 2.0 m/s can be determined using the equation:

a = (E - Bvd)/m

Plugging in the given values, we get:

a = (12 - 2.4 * 2.0 * 0.36) / 0.90 = 6.00 m/s^2

(c) The bar's terminal speed can be determined using the equation:

v_terminal = E / (Bd)

Plugging in the given values, we get:

v_terminal = 12 / (2.4 * 0.36) = 13.33 m/s