Use the worked example above to help you solve this problem. (a) The sliding bar

Question

Use the worked example above to help you solve this problem.

(a) The sliding bar in the figure has a length of 0.538 m and moves at 1.99 m/s in a magnetic field of magnitude 0.272 T. Using the concept of motional emf, find the induced voltage in the moving rod.
V

(b) If the resistance in the circuit is 0.651 ?, find the current in the circuit and the power delivered to the resistor. (Note: The current, in this case, goes counterclockwise around the loop.)

(c) Calculate the magnetic force on the bar.

(d) Use the concepts of work and power to calculate the applied force.
N

EXERCISE : Help on both plz plz

Suppose the current suddenly increases to 1.36 A in the same direction as before, due to an increase in speed of the bar. (Use the information from the PRACTICE IT part.)

I = A P = W points ! Prevos Answers SerCP11 20 AE004 EXAMPLE 20.4Where Is the Energy Source? GOAL Use motional emf to find an induced emf and a current The magnetic force F opposes the motion, and a counterclockwise current is induced in the loop PROBLEM () The sliding bar in the figure has a length of 0.500 m and moves at 2.00 m/s in a magnetic field of magnitude 0.250 T. Using the cencept of motional emf, find the induced voltage in the moving foc. (b) If the fesistance in the circuit is 0.500 ?, find the current in the circuit and the power delivered to the resistor. (Noce: The current in this case goes counterclockwise around the loop.) (e) Calculate the magnetic force on the bar. (d) Use the concepts of work and power to calculate the applied force A48 STRATEGY For part (a), substitute into |84* - Btv for the motional emf. Once the emf A2 is found, substitution into Ohm's law gives the current. In part (c), use F Basin ? for the magnetic force on a current-tarrying conductor. In part (c) use the fact that the power dissipated by the resistor multiplied by the elapsed time must equal the work done by the applied force. SOLUTION (a) Find the induced emf with the concept of motional emf Substitute to find the induced emf. -(0.250 To.500 m)(2.00 m/s) 0.250 (b)Find the induced current in the circuit and the power dissipated by the resistor. Substtute the emf and the resistance into Ohm's law to find the induced current ? 0.250,--0.500 A e-IAV-(0.500 AX0.250 V)-0.125 W Substitute0.5OD A and ? 0.250 V to find the power dissipated by the 0.500-? resistor. (c) Calculate the magnitude and direction of the maçnetic force on the bar m1B-(0.500 A)O.250 TX0.500 m) Substitute values for I, B, and into the equation, with sin ? sin(90*)-1, to find the magnitude of the force 6.25 x 102N Point the fingers of your right hand in the direction of the positive current, hen curl them in the direction of the magnetic field. Your thumb points in the negative x-direction. Apply right-hand rule number 1 to find the direction of the force. (4) Find the value of Fapp, the applied force Set the work done by the applied force ecual to the dissipated power times the elapsed time. Solve for Fao and substituted # v2c PA P P 0.125 W d 2.00 ms 6.25x 102N LEARN MORE REMARKS Part (c) could be solved by using Newton's second law for an object in equilibrium: Two forces act horizontally on the bar and the atceleration of the bar is zero, so the forces must be equal in magnitude and opposite in direction. Notice the agreement between the answers for Fm and Fapo despite the very different concepts used QUESTION Suppose the applied force and maçnetic fiek in the figure are removed, but a battery creates a current in the same direction as indicated. What happens to the bar? No magnetic force acts on it A magnetic force acts to push it away from the resistor. A magnetic force acts to push it toward the resistor O

Explanation / Answer

Part (a)

given that:-

l = 0.538 m

v = 1.99 m/s

B = 0.272 T

then the induced emf is

Vemf = Blv

Vemf = 0.272 * 0.538 * 1.99

Vemf = 0.2912 V

Part (b)

current = voltage / resistance = 0.2912 / 0.651

Current = 0.4473 A

Power = current2*Resistance

Power = 0.44732*0.651

Power = 0.13 Watt

Part (c)

Magnetic force = Magnetic field*Current*Length = 0.272*0.4473*0.538

Magnetic force = 0.0655 N

Direction by fleming's right hand rule is Negative-x axis (i.e., -X)

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

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