The current in the metal rod.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 20, Problem 3P

(a)

To determine

The current in the metal rod.

(a)

Expert Solution

Answer to Problem 3P

The current in the metal rod is 3.3×10−5A_.

Explanation of Solution

PHYSICS, Chapter 20, Problem 3P

Figure 1

Write the expression for the induced emf in the metal rod.

ε=vBL (I)

Here, ε is the induced emf, L is the length of the rod, v is the velocity.

Write the expression for the current in the metal rod.

I=εR (II)

Here, I is the current, R is the resistance.

Use equation (II) in (I) to solve for I.

I=vBLR (III)

Conclusion:

Substitute 16 cm/s for v, 0.75 T for B, 5.0 cm for L, 180 Ω for R in equation (III) to find I.

I=(16 cm×10−2m1 cm/s)(0.75 T)(5.0 cm×10−2m1 cm)(180 Ω)=0.0060 V180 Ω=3.3×10−5A

Therefore, the current in the metal rod is 3.3×10−5A_.

(b)

To determine

The energy dissipated in the resistor.

(b)

Expert Solution

Answer to Problem 3P

The rate of energy dissipated in the resistor is 2.0×10−7W_.

Explanation of Solution

Write the expression for the energy dissipated in the resistor.

P=I2R (IV)

Here, P is the power or the rate of energy dissipated in the resistor.

Conclusion:

Substitute 3.3×10−5A for I, 180 Ω for R in equation (IV) to find P.

P=(3.3×10−5A)2(180 Ω)=2.0×10−7W

Therefore, the energy dissipated in the resistor is 2.0×10−7W_.

(c)

To determine

The magnetic force on the metal rod.

(c)

Expert Solution

Answer to Problem 3P

The magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Explanation of Solution

Write the expression for the magnetic force.

F→=IL→×B→ (V)

Here, F is the magnetic force.

Conclusion:

Substitute 3.3×10−5A for I, 5.0 cm j^ for L→, 0.75 T (−k^) for B→ in equation (V) to find F→.

F→=(3.3×10−5A)(5.0 cm j^)×(0.75 T (−k^))=(3.3×10−5A)(5.0 cm)(0.75 T)(j^×−k^)=1.3×106N (−i^)

Therefore, the magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

! Required information The block shown is made of a magnesium alloy, for which E = 45 GPa and v = 0.35. Know that σx = -185 MPa. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 25 mm B D 40 mm 100 mm Determine the magnitude of Oy for which the change in the height of the block will be zero. The magnitude of Oy is MPa.

The rigid bar ABC is supported by two links, AD and BE, of uniform 37.5 × 6-mm rectangular cross section and made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σy= 250 MPa. The magnitude of the force Q applied at B is gradually increased from zero to 265 kN and a = 0.640 m. 1.7 m 1 m D A B 2.64 m E Determine the value of the normal stress in each link. The value of the normal stress in link AD is The value of the normal stress in link BE is 250 MPa. MPa.

Two tempered-steel bars, each 16 in. thick, are bonded to a ½ -in. mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude P. Both steels are elastoplastic with E= 29 × 106 psi and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered and mild steel. The load P is gradually increased from zero until the deformation of the bar reaches a maximum value dm = 0.04 in. and then decreased back to zero. Take L = 15 in. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 2.0 in. in. 3 in. 3 16 in. Determine the maximum stress in the tempered-steel bars. The maximum stress in the tempered-steel bars is ksi.

Answer 2

Question

Chapter 20, Problem 3P

(a)

To determine

The current in the metal rod.

(a)

Expert Solution

Answer to Problem 3P

The current in the metal rod is 3.3×10−5A_.

Explanation of Solution

PHYSICS, Chapter 20, Problem 3P

Figure 1

Write the expression for the induced emf in the metal rod.

ε=vBL (I)

Here, ε is the induced emf, L is the length of the rod, v is the velocity.

Write the expression for the current in the metal rod.

I=εR (II)

Here, I is the current, R is the resistance.

Use equation (II) in (I) to solve for I.

I=vBLR (III)

Conclusion:

Substitute 16 cm/s for v, 0.75 T for B, 5.0 cm for L, 180 Ω for R in equation (III) to find I.

I=(16 cm×10−2m1 cm/s)(0.75 T)(5.0 cm×10−2m1 cm)(180 Ω)=0.0060 V180 Ω=3.3×10−5A

Therefore, the current in the metal rod is 3.3×10−5A_.

(b)

To determine

The energy dissipated in the resistor.

(b)

Expert Solution

Answer to Problem 3P

The rate of energy dissipated in the resistor is 2.0×10−7W_.

Explanation of Solution

Write the expression for the energy dissipated in the resistor.

P=I2R (IV)

Here, P is the power or the rate of energy dissipated in the resistor.

Conclusion:

Substitute 3.3×10−5A for I, 180 Ω for R in equation (IV) to find P.

P=(3.3×10−5A)2(180 Ω)=2.0×10−7W

Therefore, the energy dissipated in the resistor is 2.0×10−7W_.

(c)

To determine

The magnetic force on the metal rod.

(c)

Expert Solution

Answer to Problem 3P

The magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Explanation of Solution

Write the expression for the magnetic force.

F→=IL→×B→ (V)

Here, F is the magnetic force.

Conclusion:

Substitute 3.3×10−5A for I, 5.0 cm j^ for L→, 0.75 T (−k^) for B→ in equation (V) to find F→.

F→=(3.3×10−5A)(5.0 cm j^)×(0.75 T (−k^))=(3.3×10−5A)(5.0 cm)(0.75 T)(j^×−k^)=1.3×106N (−i^)

Therefore, the magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 20, Problem 3P

(a)

To determine

The current in the metal rod.

(a)

Expert Solution

Answer to Problem 3P

The current in the metal rod is 3.3×10−5A_.

Explanation of Solution

PHYSICS, Chapter 20, Problem 3P

Figure 1

Write the expression for the induced emf in the metal rod.

ε=vBL (I)

Here, ε is the induced emf, L is the length of the rod, v is the velocity.

Write the expression for the current in the metal rod.

I=εR (II)

Here, I is the current, R is the resistance.

Use equation (II) in (I) to solve for I.

I=vBLR (III)

Conclusion:

Substitute 16 cm/s for v, 0.75 T for B, 5.0 cm for L, 180 Ω for R in equation (III) to find I.

I=(16 cm×10−2m1 cm/s)(0.75 T)(5.0 cm×10−2m1 cm)(180 Ω)=0.0060 V180 Ω=3.3×10−5A

Therefore, the current in the metal rod is 3.3×10−5A_.

(b)

To determine

The energy dissipated in the resistor.

(b)

Expert Solution

Answer to Problem 3P

The rate of energy dissipated in the resistor is 2.0×10−7W_.

Explanation of Solution

Write the expression for the energy dissipated in the resistor.

P=I2R (IV)

Here, P is the power or the rate of energy dissipated in the resistor.

Conclusion:

Substitute 3.3×10−5A for I, 180 Ω for R in equation (IV) to find P.

P=(3.3×10−5A)2(180 Ω)=2.0×10−7W

Therefore, the energy dissipated in the resistor is 2.0×10−7W_.

(c)

To determine

The magnetic force on the metal rod.

(c)

Expert Solution

Answer to Problem 3P

The magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Explanation of Solution

Write the expression for the magnetic force.

F→=IL→×B→ (V)

Here, F is the magnetic force.

Conclusion:

Substitute 3.3×10−5A for I, 5.0 cm j^ for L→, 0.75 T (−k^) for B→ in equation (V) to find F→.

F→=(3.3×10−5A)(5.0 cm j^)×(0.75 T (−k^))=(3.3×10−5A)(5.0 cm)(0.75 T)(j^×−k^)=1.3×106N (−i^)

Therefore, the magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 20, Problem 3P

(a)

To determine

The current in the metal rod.

(a)

Expert Solution

Answer to Problem 3P

The current in the metal rod is 3.3×10−5A_.

Explanation of Solution

PHYSICS, Chapter 20, Problem 3P

Figure 1

Write the expression for the induced emf in the metal rod.

ε=vBL (I)

Here, ε is the induced emf, L is the length of the rod, v is the velocity.

Write the expression for the current in the metal rod.

I=εR (II)

Here, I is the current, R is the resistance.

Use equation (II) in (I) to solve for I.

I=vBLR (III)

Conclusion:

Substitute 16 cm/s for v, 0.75 T for B, 5.0 cm for L, 180 Ω for R in equation (III) to find I.

I=(16 cm×10−2m1 cm/s)(0.75 T)(5.0 cm×10−2m1 cm)(180 Ω)=0.0060 V180 Ω=3.3×10−5A

Therefore, the current in the metal rod is 3.3×10−5A_.

(b)

To determine

The energy dissipated in the resistor.

(b)

Expert Solution

Answer to Problem 3P

The rate of energy dissipated in the resistor is 2.0×10−7W_.

Explanation of Solution

Write the expression for the energy dissipated in the resistor.

P=I2R (IV)

Here, P is the power or the rate of energy dissipated in the resistor.

Conclusion:

Substitute 3.3×10−5A for I, 180 Ω for R in equation (IV) to find P.

P=(3.3×10−5A)2(180 Ω)=2.0×10−7W

Therefore, the energy dissipated in the resistor is 2.0×10−7W_.

(c)

To determine

The magnetic force on the metal rod.

(c)

Expert Solution

Answer to Problem 3P

The magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Explanation of Solution

Write the expression for the magnetic force.

F→=IL→×B→ (V)

Here, F is the magnetic force.

Conclusion:

Substitute 3.3×10−5A for I, 5.0 cm j^ for L→, 0.75 T (−k^) for B→ in equation (V) to find F→.

F→=(3.3×10−5A)(5.0 cm j^)×(0.75 T (−k^))=(3.3×10−5A)(5.0 cm)(0.75 T)(j^×−k^)=1.3×106N (−i^)

Therefore, the magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 20, Problem 3P

(a)

To determine

The current in the metal rod.

(a)

Expert Solution

Answer to Problem 3P

The current in the metal rod is 3.3×10−5A_.

Explanation of Solution

PHYSICS, Chapter 20, Problem 3P

Figure 1

Write the expression for the induced emf in the metal rod.

ε=vBL (I)

Here, ε is the induced emf, L is the length of the rod, v is the velocity.

Write the expression for the current in the metal rod.

I=εR (II)

Here, I is the current, R is the resistance.

Use equation (II) in (I) to solve for I.

I=vBLR (III)

Conclusion:

Substitute 16 cm/s for v, 0.75 T for B, 5.0 cm for L, 180 Ω for R in equation (III) to find I.

I=(16 cm×10−2m1 cm/s)(0.75 T)(5.0 cm×10−2m1 cm)(180 Ω)=0.0060 V180 Ω=3.3×10−5A

Therefore, the current in the metal rod is 3.3×10−5A_.

(b)

To determine

The energy dissipated in the resistor.

(b)

Expert Solution

Answer to Problem 3P

The rate of energy dissipated in the resistor is 2.0×10−7W_.

Explanation of Solution

Write the expression for the energy dissipated in the resistor.

P=I2R (IV)

Here, P is the power or the rate of energy dissipated in the resistor.

Conclusion:

Substitute 3.3×10−5A for I, 180 Ω for R in equation (IV) to find P.

P=(3.3×10−5A)2(180 Ω)=2.0×10−7W

Therefore, the energy dissipated in the resistor is 2.0×10−7W_.

(c)

To determine

The magnetic force on the metal rod.

(c)

Expert Solution

Answer to Problem 3P

The magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Explanation of Solution

Write the expression for the magnetic force.

F→=IL→×B→ (V)

Here, F is the magnetic force.

Conclusion:

Substitute 3.3×10−5A for I, 5.0 cm j^ for L→, 0.75 T (−k^) for B→ in equation (V) to find F→.

F→=(3.3×10−5A)(5.0 cm j^)×(0.75 T (−k^))=(3.3×10−5A)(5.0 cm)(0.75 T)(j^×−k^)=1.3×106N (−i^)

Therefore, the magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Answer 6

Chapter 20, Problem 3P

(a)

To determine

The current in the metal rod.

(a)

Expert Solution

Answer to Problem 3P

The current in the metal rod is 3.3×10−5A_.

Explanation of Solution

PHYSICS, Chapter 20, Problem 3P

Figure 1

Write the expression for the induced emf in the metal rod.

ε=vBL (I)

Here, ε is the induced emf, L is the length of the rod, v is the velocity.

Write the expression for the current in the metal rod.

I=εR (II)

Here, I is the current, R is the resistance.

Use equation (II) in (I) to solve for I.

I=vBLR (III)

Conclusion:

Substitute 16 cm/s for v, 0.75 T for B, 5.0 cm for L, 180 Ω for R in equation (III) to find I.

I=(16 cm×10−2m1 cm/s)(0.75 T)(5.0 cm×10−2m1 cm)(180 Ω)=0.0060 V180 Ω=3.3×10−5A

Therefore, the current in the metal rod is 3.3×10−5A_.

Answer 7

Chapter 20, Problem 3P

(a)

To determine

The current in the metal rod.

Answer 8

(a)

Expert Solution

Answer to Problem 3P

The current in the metal rod is 3.3×10−5A_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

PHYSICS, Chapter 20, Problem 3P

Figure 1

Write the expression for the induced emf in the metal rod.

ε=vBL (I)

Here, ε is the induced emf, L is the length of the rod, v is the velocity.

Write the expression for the current in the metal rod.

I=εR (II)

Here, I is the current, R is the resistance.

Use equation (II) in (I) to solve for I.

I=vBLR (III)

Conclusion:

Substitute 16 cm/s for v, 0.75 T for B, 5.0 cm for L, 180 Ω for R in equation (III) to find I.

I=(16 cm×10−2m1 cm/s)(0.75 T)(5.0 cm×10−2m1 cm)(180 Ω)=0.0060 V180 Ω=3.3×10−5A

Therefore, the current in the metal rod is 3.3×10−5A_.

Answer 13

(b)

To determine

The energy dissipated in the resistor.

(b)

Expert Solution

Answer to Problem 3P

The rate of energy dissipated in the resistor is 2.0×10−7W_.

Explanation of Solution

Write the expression for the energy dissipated in the resistor.

P=I2R (IV)

Here, P is the power or the rate of energy dissipated in the resistor.

Conclusion:

Substitute 3.3×10−5A for I, 180 Ω for R in equation (IV) to find P.

P=(3.3×10−5A)2(180 Ω)=2.0×10−7W

Therefore, the energy dissipated in the resistor is 2.0×10−7W_.

Answer 14

(b)

To determine

The energy dissipated in the resistor.

Answer 15

(b)

Expert Solution

Answer to Problem 3P

The rate of energy dissipated in the resistor is 2.0×10−7W_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for the energy dissipated in the resistor.

P=I2R (IV)

Here, P is the power or the rate of energy dissipated in the resistor.

Conclusion:

Substitute 3.3×10−5A for I, 180 Ω for R in equation (IV) to find P.

P=(3.3×10−5A)2(180 Ω)=2.0×10−7W

Therefore, the energy dissipated in the resistor is 2.0×10−7W_.

Answer 20

(c)

To determine

The magnetic force on the metal rod.

(c)

Expert Solution

Answer to Problem 3P

The magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Explanation of Solution

Write the expression for the magnetic force.

F→=IL→×B→ (V)

Here, F is the magnetic force.

Conclusion:

Substitute 3.3×10−5A for I, 5.0 cm j^ for L→, 0.75 T (−k^) for B→ in equation (V) to find F→.

F→=(3.3×10−5A)(5.0 cm j^)×(0.75 T (−k^))=(3.3×10−5A)(5.0 cm)(0.75 T)(j^×−k^)=1.3×106N (−i^)

Therefore, the magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Answer 21

(c)

To determine

The magnetic force on the metal rod.

Answer 22

(c)

Expert Solution

Answer to Problem 3P

The magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression for the magnetic force.

F→=IL→×B→ (V)

Here, F is the magnetic force.

Conclusion:

Substitute 3.3×10−5A for I, 5.0 cm j^ for L→, 0.75 T (−k^) for B→ in equation (V) to find F→.

F→=(3.3×10−5A)(5.0 cm j^)×(0.75 T (−k^))=(3.3×10−5A)(5.0 cm)(0.75 T)(j^×−k^)=1.3×106N (−i^)

Therefore, the magnetic force on the metal rod is 1.3×10−6N in the left direction_.

Answer 27

Ch. 20.1 - 20.1 If the rod in Fig. 20.1 were moving out of...Ch. 20.1 - Conceptual Practice Problem 20.1 Loop of Different...Ch. 20.2 - Prob. 20.2PP Ch. 20.3 - Prob. 20.3PP Ch. 20.3 - Practice Problem 20.4 Rotating Coil Generator In a...Ch. 20.4 - Figure 20.11 Circular loop in a magnetic field of...Ch. 20.4 - Prob. 20.5PP Ch. 20.4 - Prob. 20.6PP Ch. 20.6 - CHECKPOINT 20.6 The primary coil of a...Ch. 20.6 - Practice Problem 20.7 An Ideal Transformer An...

The current in the metal rod.

Concept explainers

The current in the metal rod.

Answer to Problem 3P

Explanation of Solution

The energy dissipated in the resistor.

Answer to Problem 3P

Explanation of Solution

The magnetic force on the metal rod.

Answer to Problem 3P

Explanation of Solution

Want to see more full solutions like this?

Chapter 20 Solutions