College Physics:
College Physics:
11th Edition
ISBN: 9781305965515
Author: SERWAY, Raymond A.
Publisher: Brooks/Cole Pub Co
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Chapter 20, Problem 30P

(a)

To determine

The constant speed (v) should the bar be moved to produce a current (I) of 1.00A in the resistor

(a)

Expert Solution
Check Mark

Answer to Problem 30P

Solution: The constant speed (v) should the bar be moved to produce a current (I) of 1.00A in the resistor is 2.00m/s

Explanation of Solution

Given Info: Resistance R is 6.00Ω , length l is 1.20m , current I is 1.00A and magnetic field B is 2.50T ,

The motional emf is

ΔV=Blv

  • ΔV is the motional emf,
  • B is the magnetic field,
  • l is the length of the bar,
  • v is the speed of the bar,

Formula to calculate thecurrent is,

I=ΔVR

  • I is the current in the rod,
  • R is the resistance of the rod,

Use Blv for ΔV in the above equation to find the expression for I

I=BlvR

v=IRBl

Substitute 6.00Ω for R, 1.00A for I, 2.50T for Band 1.20m for lin the above equation to find I

I=(6.00Ω)(1.00A)(2.50T)(1.20m)=2.00m/s

Conclusion:

The constant speed (v) should the bar be moved to produce a current (I) of 1.00A in the resistor is 2.00m/s

(b)

To determine

The power delivered to the resistor

(b)

Expert Solution
Check Mark

Answer to Problem 30P

Solution: The power delivered to the resistor P is 6.00W

Explanation of Solution

Given Info: Resistance R is 6.00Ω , Current I is 1.00A .

Formula to calculate Power delivered to the resistor is,

P=I2R

  • P is the power delivered to the resistor,

Substitute 6.00Ω for R and 1.00A for I to find P.

P=(1.00A)2(6.00Ω)=6.00W

Conclusion:

The power delivered to the resistor P is 6.00W

(c)

To determine

The magnetic force (F) is exerted on the moving bar.

(c)

Expert Solution
Check Mark

Answer to Problem 30P

Solution: The magnetic force (F) is exerted on the moving bar is 3.00N .

Explanation of Solution

Given Info: Length l is 1.20m , angle θ is 90° current I is 1.00A , and magnetic field B is 2.50T

Formula to calculate magnetic force (F) is exerted on the moving bar.

F=Blvsinθ

  • F is the magnetic force on the bar,

Substitute, 1.00A for I, 2.50T for B, 90° for θ , and 1.20m for l in the above equation to find I .

F=(2.50T)(1.00A)(1.20m)(sin90°)=3.00N

Conclusion:

The magnetic force (F) is exerted on the moving bar is 3.00N .

(d):

To determine

The power delivered by the force Fapp on the moving bar.

(d):

Expert Solution
Check Mark

Answer to Problem 30P

Solution: The power delivered by the force Fapp on the moving bar is 6.00W

Explanation of Solution

Given Info: speed vis 2.00m/s and Fapp is 3.00N

The bar moves at the constant speed in the same direction. So, the acceleration is a=0

F=Fapp

Formula to calculate power delivered by the force Fapp on the moving bar is

P=F.v

Use Fapp for F in the above equation to find the expression for P.

P=Fapp.v

  • P is the power delivered.
  • Fapp is the force applied at instant

Substitute 3.00N for Fapp and 2.00m/s for vto find P

P=(3.00N)(2.00m/s)=6.00W

Conclusion:

The power delivered by the force Fapp on the moving bar is 6.00W

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Chapter 20 Solutions

College Physics:

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