A current- carrying straight wire is kept along the axis of a circular loop carrying a current. The straight wire

Options

A. will exert an inward force on the circular loop
B. will exert an outward force on the circular loop
C. will not exert any force on the circular loop
D. will exert a force on the circular loop parallel to itself.

Answer

To solve the problem of a current-carrying straight wire kept along the axis of a circular loop carrying a current, we will analyze the magnetic fields and forces involved.

### Step-by-Step Solution:

1. **Identify the Setup**:
- We have a circular loop carrying a current (let's denote this current as \( I_1 \)).
- A straight wire carrying a current (denote this as \( I_2 \)) is placed along the axis of the circular loop.

2. **Determine the Direction of Current**:
- Assume the current in the circular loop flows in a clockwise direction when viewed from one end.
- The straight wire also carries a current, which we will assume flows in a specific direction (let's say upward for this explanation).

3. **Use the Right-Hand Rule**:
- For the straight wire, apply the right-hand rule to determine the direction of the magnetic field it produces.
- Point your thumb in the direction of the current in the straight wire; your fingers will curl in the direction of the magnetic field lines.
- The magnetic field due to the straight wire will form concentric circles around the wire.

4. **Magnetic Field at the Loop**:
- At the center of the circular loop, the magnetic field \( B \) due to the straight wire can be calculated using the formula:
\[ B = \frac{\mu_0 I_2}{2\pi r} \]
where \( r \) is the distance from the wire to the center of the loop.

5. **Force on the Circular Loop**:
- The force \( F \) on a segment of the circular loop due to the magnetic field from the straight wire can be calculated using the formula:
\[ F = I_1 \, dl \times B \]
- Here, \( dl \) is the differential length element of the loop.

6. **Direction of Force**:
- Since the magnetic field \( B \) and the current \( I_1 \) in the loop are perpendicular to each other at the center of the loop, the angle between \( dl \) and \( B \) is 0 degrees.
- Therefore, \( F = I_1 \, dl \, B \sin(0) = 0 \).
- This indicates that there is no net force acting on the circular loop due to the straight wire.

7. **Conclusion**:
- Since the force on the circular loop is zero, it will remain in equilibrium and will not experience any movement due to the presence of the straight wire.

### Final Answer:
The correct option is that no force is exerted on the circular loop due to the straight wire.