23 of 23 Constants Part A A 2.60-N metal bar, 0.850 m long and having a resistan

Question

23 of 23 Constants Part A A 2.60-N metal bar, 0.850 m long and having a resistance of 10.02, rests horizontally on conducting wires connecting it to the circuit shown in (Figure 1). The bar is in a uniform, horizontal, 1.60-T magnetic field and is not attached to the wires in the circuit. What is the acceleration of the bar just after the switch S is olosed? Express your answer with the appropriate units 0.905 m Submit Previous Answers Request Answer Incorrect: Try Again: 4 attempts remaining Figure 1 of 1

Explanation / Answer

Solution :-

Req = 25 + ( 10 X 10 ) / ( 10 + 10 )

Req = 25 + 100 / 20

Req = 25 + 5

Req = 30 Ohm

the current through 25 Ohm is

i25 = 240 / 30

i25 = 4 A

now the current through the rod is

irod = i25 X 5 / 10

irod = 4 X 5 / 10

irod = 20 / 2

irod = 2 A

the force is " F "

F = irod X L X B

= 2 X 0.850 X 1.60

= 2.72 N

now the acceleration of the bar just after the switch is closed

a = F X g / m

a = 2.72 X 9.8 / 2.6

a = 10.2523 m/sec2

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.