A metal rod of resistance `20 Omega` is fixed along a diameter of a conducting ring of radius `0.1 m` and lies on `x-y` plane. There is a magnetic field `vec(B) = (50 T)vec(k)`. The ring rotates with an angular velocity `omega = 20rad s^(-1)` about its axis. An external resistance of `10 Omega` is connected across the center of the ring and rim. The current external resistance is

Options

A. `(1)/(4)A`
B. `(1)/(2)A`
C. `(1)/(3)A`
D. ZERO

Answer

To find the current in the external resistance connected across the conducting ring, we can follow these steps:

### Step 1: Calculate the induced EMF (Electromotive Force)
The induced EMF (E) in the rotating ring can be calculated using the formula:
\[ E = \frac{1}{2} B \omega r^2 \]
Where:
- \( B = 50 \, T \) (magnetic field)
- \( \omega = 20 \, \text{rad/s} \) (angular velocity)
- \( r = 0.1 \, m \) (radius of the ring)

Substituting the values:
\[ E = \frac{1}{2} \times 50 \times 20 \times (0.1)^2 \]
\[ E = \frac{1}{2} \times 50 \times 20 \times 0.01 \]
\[ E = \frac{1}{2} \times 50 \times 20 \times 0.01 = \frac{500}{100} = 5 \, V \]

### Step 2: Analyze the circuit configuration
The metal rod of resistance \( 20 \, \Omega \) is fixed along the diameter of the ring. This means that the rod can be considered as two resistors of \( 10 \, \Omega \) each in parallel since the total resistance is \( 20 \, \Omega \).

### Step 3: Determine the equivalent resistance of the rod
Since the two halves of the rod are in parallel:
\[ R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{10 \times 10}{10 + 10} = \frac{100}{20} = 5 \, \Omega \]

### Step 4: Total resistance in the circuit
The total resistance in the circuit will be the sum of the equivalent resistance of the rod and the external resistance:
\[ R_{total} = R_{eq} + R_{external} = 5 \, \Omega + 10 \, \Omega = 15 \, \Omega \]

### Step 5: Calculate the total current in the circuit
Using Ohm's law, the current \( I \) can be calculated as:
\[ I = \frac{E}{R_{total}} = \frac{5 \, V}{15 \, \Omega} = \frac{1}{3} \, A \]

### Final Answer
The current in the external resistance is:
\[ I = \frac{1}{3} \, A \]

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