Hook a sista up please. Explanations are awesome! Problem 20.87 A 2.85-N metal b

Question

Hook a sista up please. Explanations are awesome!

Problem 20.87 A 2.85-N metal bar. 1.20m long and having a resistance of 10.0 ohm. rests horizontally on conducting wires connecting it to the circuit shown in the figure (Figure 1). The bar is in a uniform. horizontal. 1.80-T magnetic field and is not attached to the wires in the circuit. Part A What is the magnitude of the acceleration of the bar just after the switch S is closed? Part B What is the direction of the acceleration of the bar just after the switch S is closed? To the left Upward To the right Downward

Explanation / Answer

Note that the bar is parallel to the 10 ohm resistor.

Thus, using parallel resistors, their net resistance is

Rnet = 5 ohms

Now, we have a 25 ohm and 5 ohm resistor in series.

Thus, the current that goes to the 5 ohm resistor is

I = V/(R1 + R2)

I = 120/(25 + 5)

I = 4.8 A

So, 4.8 A is distributed to the two 10 ohm resistors. Thus, the current on the bar is half of this,

I_bar = 2.4 A

Note that

F = BIL

Thus, as B = 1.80 T, L = 1.20 m,

F = 5.184 N

Thus, the acceleration, a = F/m, as m = 2.85/9.8 kg = 0.2908 kg is

a = 5.184 N/0.2908 kg

a = 17.8 m/s^2 [ANSWER]

**********************************

Current flows from the + side to the - side.

Thus, the current on B is downward.

By right hand rule, the force will be upward.

Thus, the acceleration is also UPWARD. [ANSWER]

DONE! It was nice working with you!

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