Two long parallel conducting rails are separated by a distance d. At one end they are joined by a resistance R. A conducting bar of length d is made to slide at constant velocity v along the rails. Both the bar and the rails have negligible resistance. (a) What current flows in the circuit? (b) How much power is required to move the bar? (c) How does the power dissipated in the resistance compare with the power required to move the bar?

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### Physics Problem: Conducting Rails and Moving Bar

#### Problem Description
Two long parallel conducting rails are separated by a distance \( d \). At one end, they are joined by a resistance \( R \). A conducting bar of length \( d \) is made to slide at constant velocity \( v \) along the rails. Both the bar and the rails have negligible resistance.

**Questions:**
1. **(a)** What current flows in the circuit?
2. **(b)** How much power is required to move the bar?
3. **(c)** How does the power dissipated in the resistance compare with the power required to move the bar?

---

To solve this problem, it is essential to analyze the circuit and apply relevant electromagnetic theories.

### Detailed Explanation:

1. **Current Flow in the Circuit (Part a):**
    - The sliding bar induces an electromotive force (emf) due to its motion in a magnetic field. By Faraday's Law of Induction, the emf (\( \epsilon \)) generated in the bar moving perpendicularly through a magnetic field \( B \) with a constant velocity \( v \) is given by:
      \[
      \epsilon = B \cdot v \cdot d
      \]
    - According to Ohm's law, the current \( I \) flowing through the circuit, where the resistance is \( R \), can be calculated as:
      \[
      I = \frac{\epsilon}{R} = \frac{B \cdot v \cdot d}{R}
      \]

2. **Power Required to Move the Bar (Part b):**
    - The power \( P \) required to move the bar at a constant velocity against the induced magnetic force involves calculating the force \( F \) exerted on the bar due to the current and then finding the power required to overcome this force. The magnetic force is given by:
      \[
      F = I \cdot d \cdot B
      \]
    - The power required to move the bar with this force at velocity \( v \) is:
      \[
      P = F \cdot v = (I \cdot d \cdot B) \cdot v
      \]
    - Substituting the value of \( I \) from above:
      \[
      P = \left( \frac{B \cdot v \cdot
Transcribed Image Text:### Physics Problem: Conducting Rails and Moving Bar #### Problem Description Two long parallel conducting rails are separated by a distance \( d \). At one end, they are joined by a resistance \( R \). A conducting bar of length \( d \) is made to slide at constant velocity \( v \) along the rails. Both the bar and the rails have negligible resistance. **Questions:** 1. **(a)** What current flows in the circuit? 2. **(b)** How much power is required to move the bar? 3. **(c)** How does the power dissipated in the resistance compare with the power required to move the bar? --- To solve this problem, it is essential to analyze the circuit and apply relevant electromagnetic theories. ### Detailed Explanation: 1. **Current Flow in the Circuit (Part a):** - The sliding bar induces an electromotive force (emf) due to its motion in a magnetic field. By Faraday's Law of Induction, the emf (\( \epsilon \)) generated in the bar moving perpendicularly through a magnetic field \( B \) with a constant velocity \( v \) is given by: \[ \epsilon = B \cdot v \cdot d \] - According to Ohm's law, the current \( I \) flowing through the circuit, where the resistance is \( R \), can be calculated as: \[ I = \frac{\epsilon}{R} = \frac{B \cdot v \cdot d}{R} \] 2. **Power Required to Move the Bar (Part b):** - The power \( P \) required to move the bar at a constant velocity against the induced magnetic force involves calculating the force \( F \) exerted on the bar due to the current and then finding the power required to overcome this force. The magnetic force is given by: \[ F = I \cdot d \cdot B \] - The power required to move the bar with this force at velocity \( v \) is: \[ P = F \cdot v = (I \cdot d \cdot B) \cdot v \] - Substituting the value of \( I \) from above: \[ P = \left( \frac{B \cdot v \cdot
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