A metal rod of resistance `R` is fixed along a diameter o fa conducting ring of radius `r` . There is a magnetic field of magnitude `B` perpendicular to the plane of the loop. The ring spins with an angular velocity `omega` about its axis. The centre of the ring is joined to its rim by an external wire `W` . The ring and `W` have no resistance. The current in `W` is

Options

A. zero
B. `(Br^(2)omega)/(2R)`
C. `(Br^(2)omega)/(R)`
D. `(2Br^(2)omega)/(R)`

Explanation

To solve the problem, we need to analyze the situation step by step, focusing on the induced electromotive force (EMF) and the resulting current in the wire `W`.

### Step 1: Understand the Setup
We have a conducting ring of radius `r` with a metal rod of resistance `R` fixed along its diameter. The ring is spinning with an angular velocity `ω` in a magnetic field `B` that is perpendicular to the plane of the ring.

### Step 2: Determine the Induced EMF
When the ring rotates in the magnetic field, an electromotive force (EMF) is induced across the diameter of the rod. The formula for the induced EMF (E) in a rotating conductor in a magnetic field is given by:
\[ E = B \cdot v \cdot L \]
where:
- \( v \) is the linear velocity of the point on the ring,
- \( L \) is the length of the conductor in the magnetic field.

For a point on the rim of the ring, the linear velocity \( v \) can be expressed as:
\[ v = r \cdot \omega \]
Thus, the EMF across the rod (which has a length equal to the diameter of the ring) becomes:
\[ E = B \cdot (r \cdot \omega) \cdot (2r) = 2Br^2\omega \]

### Step 3: Analyze the Circuit
The rod divides the ring into two halves, and since the rod has resistance `R`, we can consider each half of the rod to have a resistance of \( \frac{R}{2} \). The wire `W` connecting the center to the rim has no resistance.

### Step 4: Set Up the Equivalent Circuit
The two halves of the rod can be treated as two batteries (due to the induced EMF) in parallel:
- Each half has an EMF of \( E \) and resistance \( \frac{R}{2} \).

The equivalent EMF \( E_{net} \) of the two halves in parallel is:
\[ E_{net} = E \]
The equivalent resistance \( R_{net} \) of the two halves in parallel is given by:
\[ R_{net} = \frac{R/2 \cdot R/2}{R/2 + R/2} = \frac{R/4} \]

### Step 5: Calculate the Current in Wire W
Using Ohm's law, the current \( I \) flowing through the wire `W` can be calculated as:
\[ I = \frac{E_{net}}{R_{net}} \]

Substituting the values we found:
\[ I = \frac{2Br^2\omega}{R/4} = \frac{8Br^2\omega}{R} \]

### Final Answer
The current in wire `W` is:
\[ I = \frac{8Br^2\omega}{R} \]

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