A battery is connected between two points `A and B` on the circumference of a uniform conducting ring of radius `r` and resistance `R` . One of the arcs `AB` of the ring subtends an angle `theta` at the centre . The value of the magnetic induction at the centre due to the current in the ring is

Options

A. proportional to `2(180^(@) - theta)`
B. inversely proportional to r
C. zero, only if `theta=180^(@)`
D. zero for all values of `theta`

Answer

To find the magnetic induction at the center of a uniform conducting ring with a battery connected between two points A and B on its circumference, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a conducting ring of radius \( r \) and resistance \( R \).
- Points A and B on the circumference are connected to a battery, creating a current in the ring.
- The arc \( AB \) subtends an angle \( \theta \) at the center of the ring.

2. **Identify the Current Distribution**:
- The current \( I \) flowing through the ring can be divided into two segments:
- Segment \( I_1 \) for the arc \( AB \) (subtending angle \( \theta \)).
- Segment \( I_2 \) for the remaining arc (subtending angle \( 2\pi - \theta \)).
- The relationship between these currents can be derived from the lengths of the arcs:
\[ I_1 \propto r\theta \quad \text{and} \quad I_2 \propto r(2\pi - \theta) \]

3. **Using Ohm's Law**:
- According to Ohm's law, the current is inversely proportional to resistance. Thus, we can express the currents as:
\[ I_1 = k \cdot r\theta \quad \text{and} \quad I_2 = k \cdot r(2\pi - \theta) \]
- Here, \( k \) is a constant of proportionality.

4. **Determine the Magnetic Field Contribution**:
- The magnetic field at the center of the ring due to a current-carrying segment is given by:
\[ B = \frac{\mu_0 I}{2R} \cdot \frac{\text{Angle subtended}}{2\pi} \]
- For segment \( I_1 \):
\[ B_1 = \frac{\mu_0 I_1}{2R} \cdot \frac{\theta}{2\pi} = \frac{\mu_0 I_1 \theta}{4\pi R} \]
- For segment \( I_2 \):
\[ B_2 = \frac{\mu_0 I_2}{2R} \cdot \frac{2\pi - \theta}{2\pi} = \frac{\mu_0 I_2 (2\pi - \theta)}{4\pi R} \]

5. **Calculate the Net Magnetic Field**:
- The net magnetic field \( B_{\text{net}} \) at the center due to both segments will be:
\[ B_{\text{net}} = B_1 - B_2 \]
- Substituting the expressions for \( B_1 \) and \( B_2 \):
\[ B_{\text{net}} = \frac{\mu_0 I_1 \theta}{4\pi R} - \frac{\mu_0 I_2 (2\pi - \theta)}{4\pi R} \]

6. **Simplifying the Expression**:
- From the current relationships, we can express \( I_1 \) and \( I_2 \) in terms of \( I \):
\[ I_1 = \frac{I \cdot \theta}{2\pi} \quad \text{and} \quad I_2 = \frac{I (2\pi - \theta)}{2\pi} \]
- Substituting these into the expression for \( B_{\text{net}} \):
\[ B_{\text{net}} = \frac{\mu_0 I \theta}{4\pi R} \cdot \frac{\theta}{2\pi} - \frac{\mu_0 I (2\pi - \theta)}{4\pi R} \cdot \frac{(2\pi - \theta)}{2\pi} \]

7. **Final Result**:
- After simplifying, we find that the contributions from both segments cancel each other out, leading to:
\[ B_{\text{net}} = 0 \]
- Therefore, the magnetic induction at the center of the ring is zero for all values of \( \theta \).