The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t 1 . What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times? Figure P6.1 Loops of Problem 6.1.

Question

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Answer 1

Textbook Question

Chapter 6, Problem 1P

The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t₁. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times?

Chapter 6, Problem 1P, The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t1.

Figure P6.1 Loops of Problem 6.1.

Expert Solution & Answer

To determine

The direction of the current I in the top loop.

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.

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02:10

Students have asked these similar questions

The initial voltage on the 0.5 μF capacitor shown in Fig. P6.19(a) is −20 V. The capacitor current has the waveform shown in Fig. P6.19(b). 1. How much energy, in microjoules, is stored in the capacitor at t=500 μs? 2. Repeat (a) for t=∞.

finnd the Electric Field at P (6, -3, -4) due to: Q_{1} = - 30mu*C at (2, 3, -6)

Find the Electric Field at P(6, - 3, - 4) due to: Q_{1} = - 30mu*C at (2, 3, -6)

Answer 2

Textbook Question

Chapter 6, Problem 1P

The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t₁. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times?

Chapter 6, Problem 1P, The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t1.

Figure P6.1 Loops of Problem 6.1.

Expert Solution & Answer

To determine

The direction of the current I in the top loop.

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.

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02:10

Answer 3

Textbook Question

Chapter 6, Problem 1P

The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t₁. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times?

Chapter 6, Problem 1P, The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t1.

Figure P6.1 Loops of Problem 6.1.

Expert Solution & Answer

To determine

The direction of the current I in the top loop.

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.

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02:10

Answer 4

Textbook Question

Chapter 6, Problem 1P

The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t₁. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times?

Chapter 6, Problem 1P, The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t1.

Figure P6.1 Loops of Problem 6.1.

Expert Solution & Answer

To determine

The direction of the current I in the top loop.

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.

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02:10

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The direction of the current I in the top loop.

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.

Answer 8

Expert Solution & Answer

To determine

The direction of the current I in the top loop.

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The direction of the current I in the top loop.

Answer 12

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Answer 13

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals Of Applied Electromagnetics, Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.