Question

Let a rod \(AB\) of length \(l\) having resistance \(r\) is moving perpendicular to a magnetic field \(B\) with constant velocity \(v\). If the ends of the rod are connected to a wire \(PQRS\) of negligible resistance, then current passing through the wire will be:

• A. \(\displaystyle \frac{Blv}{2r}\)
• B. \(\displaystyle \frac{Blv}{r}\)
• C. \(\displaystyle \frac{Blv}{4}\)
• D. \(\displaystyle \frac{3Blv}{4r}\)

Hint:

For a rod moving perpendicular to a magnetic field, induced emf is \(\varepsilon=Blv\). Then use \(I=\frac{\varepsilon}{R}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When a conducting rod moves through a magnetic field, the free charges within the conductor experience a Lorentz force. This force causes the charges to accumulate at the ends of the rod, creating a potential difference called motional electromotive force (EMF).
If the rod is connected to an external circuit, this induced EMF drives an electric current through the loop.
Step 2: Key Formula or Approach:
1. The induced EMF (\(\varepsilon\)) for a rod of length \(l\) moving with velocity \(v\) perpendicular to a magnetic field \(B\) is:
\[ \varepsilon = Blv \]
2. According to Ohm's Law, the current (\(I\)) in a circuit is given by:
\[ I = \frac{\text{Total EMF}}{\text{Total Resistance}} \]
Step 3: Detailed Explanation:
In this scenario, the moving rod \(AB\) acts like a battery with an EMF of \(Blv\) and an internal resistance \(r\).
The rod is connected to a wire loop \(PQRS\). The problem states that the resistance of this external wire \(PQRS\) is negligible (effectively zero).
The total resistance (\(R_{eq}\)) of the closed circuit is:
\[ R_{eq} = \text{Resistance of rod } AB + \text{Resistance of wire } PQRS \]
\[ R_{eq} = r + 0 = r \]
The current \(I\) flowing through the circuit (and thus through the wire \(PQRS\)) is:
\[ I = \frac{\varepsilon}{R_{eq}} \]
\[ I = \frac{Blv}{r} \]
This current will flow in a direction determined by Lenz's Law or the Right-Hand Palm Rule.
Step 4: Final Answer:
The current passing through the wire is \(\frac{Blv}{r}\).