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

Concept:
When a conducting rod of length \(l\) moves with velocity \(v\) perpendicular to a magnetic field \(B\), motional emf is induced across its ends. The induced emf is: \[ \varepsilon=Blv \]

Step 1: Write the induced emf.
For the moving rod: \[ \varepsilon=Blv \]

Step 2: Find total resistance of the circuit.
The wire \(PQRS\) has negligible resistance. Only the rod has resistance \(r\). Therefore, total resistance is: \[ R=r \]

Step 3: Apply Ohm's law.
Current is: \[ I=\frac{\varepsilon}{R} \] \[ I=\frac{Blv}{r} \]

Step 4: Final conclusion.
Hence, the current passing through the wire will be: \[ \boxed{\frac{Blv}{r}} \]