A cell is connected between two points of a uniformly thick circular conductor. The magnetic field at the centre of the loop will be
(Here `i_(1)` and `i_(2)` are the currents flowing in the two parts of the circular conductor of radius `'a'` and `mu_(0)` has the usual meaning)

Options

A. Zero
B. `(mu_(0))/(2a)(i_(1)-i_(2))`
C. `(mu_(0))/(2a)(i_(1)+i_(2))`
D. `(mu_(0))/a(i_(1)+i_(2))`

Answer

To find the magnetic field at the center of a uniformly thick circular conductor with two different currents flowing in opposite directions, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a circular conductor of radius 'a'.
- There are two segments of the conductor carrying currents \(i_1\) and \(i_2\) in opposite directions.
- The magnetic field at the center of the loop needs to be calculated.

2. **Magnetic Field Due to a Circular Loop**:
- The magnetic field \(B\) at the center of a circular loop carrying current \(I\) is given by the formula:
\[ B = \frac{\mu_0 I}{2a} \]
- Here, \(\mu_0\) is the permeability of free space, and \(a\) is the radius of the loop.

3. **Calculating Magnetic Fields for Each Segment**:
- For the segment carrying current \(i_1\), the magnetic field at the center is:
\[ B_1 = \frac{\mu_0 i_1}{2a} \]
- For the segment carrying current \(i_2\), the magnetic field at the center is:
\[ B_2 = \frac{\mu_0 i_2}{2a} \]

4. **Direction of Magnetic Fields**:
- The direction of the magnetic field due to \(i_1\) can be determined using the right-hand rule. If \(i_1\) is flowing in a clockwise direction, \(B_1\) will be directed into the plane.
- Conversely, if \(i_2\) is flowing in a counterclockwise direction, \(B_2\) will be directed out of the plane.

5. **Net Magnetic Field Calculation**:
- Since \(B_1\) and \(B_2\) are in opposite directions, the net magnetic field \(B_{net}\) at the center is:
\[ B_{net} = B_1 - B_2 \]
- Substituting the expressions for \(B_1\) and \(B_2\):
\[ B_{net} = \frac{\mu_0 i_1}{2a} - \frac{\mu_0 i_2}{2a} \]
- This simplifies to:
\[ B_{net} = \frac{\mu_0}{2a} (i_1 - i_2) \]

6. **Final Result**:
- The magnetic field at the center of the loop is given by:
\[ B_{net} = \frac{\mu_0}{2a} (i_1 - i_2) \]