Suppose an isolated north pole is kept at the centre of a circular loop carrying a electric current i. The magnetic field due to the north pole at a point on the periphery of the wire is B. The radius of the loop is a . The force on the wire is

Options

A. Nearly `2 pi a I B` perpendicular to the plane of the wire
B. `2pi a I B` in the plane of the wire
C. `pi aiB` along the axis of the wire
D. zero

Answer

To solve the problem, we need to determine the force on a circular wire carrying an electric current \( I \) when an isolated north pole is placed at its center.

### Step-by-Step Solution:

1. **Identify the Setup**:
We have a circular loop of radius \( a \) carrying a current \( I \). An isolated north pole is placed at the center of this loop.

2. **Magnetic Field Due to the North Pole**:
The magnetic field \( B \) at a point on the periphery of the loop (which is at a distance \( a \) from the north pole) is given. This field is directed outwards from the north pole.

3. **Consider a Small Segment of the Wire**:
Let’s consider a small segment of the wire of length \( dl \). The force \( dF \) on this segment due to the magnetic field \( B \) can be expressed using the formula:
\[ dF = I \cdot dl \cdot B \]
where \( I \) is the current flowing through the wire, \( dl \) is the length of the wire segment, and \( B \) is the magnetic field at that point.

4. **Total Force on the Wire**:
To find the total force \( F \) on the entire wire, we need to integrate \( dF \) around the loop. The total length of the circular loop is the circumference, which is \( 2\pi a \). Therefore, we can express the total force as:
\[ F = \int dF = \int I \cdot dl \cdot B = I \cdot B \cdot \int dl \]
Since \( \int dl = 2\pi a \), we have:
\[ F = I \cdot B \cdot (2\pi a) \]

5. **Final Expression for the Force**:
Thus, the total force on the wire is:
\[ F = 2\pi a I B \]

### Conclusion:
The force on the wire due to the magnetic field created by the isolated north pole is \( F = 2\pi a I B \).