A very long bar magnet is placed with its north pole coinciding with the centre of a circular loop carrying an electric current. `i`. The magnetic field due to the magnet 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. very nearly `2piaiB` perpendicular to the plane of the wire.
B. 2piaiB` in the plane of the wire
C. `piaiB` along the magnet
D. zero

Answer

To solve the problem, we need to find the force acting on a circular loop carrying an electric current \( I \) when placed in the magnetic field \( B \) produced by a long bar magnet. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a long bar magnet with its north pole at the center of a circular loop of radius \( a \). The loop carries a current \( I \). The magnetic field \( B \) at the periphery of the loop due to the magnet is given.

### Step 2: Consider a Small Element of the Loop
Take a small segment of the loop, denoted as \( dl \). The force \( dF \) acting on this small segment due to the magnetic field can be expressed using the formula:
\[ dF = B \cdot I \cdot dl \]
where \( B \) is the magnetic field at the location of the segment, \( I \) is the current in the loop, and \( dl \) is the length of the small segment.

### Step 3: Determine the Direction of the Force
The direction of the force can be determined using the right-hand rule. The force \( dF \) acts perpendicular to both the magnetic field \( B \) and the current \( I \) in the wire.

### Step 4: Integrate Over the Entire Loop
To find the total force \( F \) on the entire loop, we need to integrate \( dF \) around the loop. Since the loop is circular, we can express the total force as:
\[ F = \int dF = \int B \cdot I \cdot dl \]
The total length of the loop is the circumference, which is \( 2\pi a \).

### Step 5: Substitute and Solve
Substituting the expression for \( dF \) into the integral:
\[ F = \int_0^{2\pi} B \cdot I \cdot dl = B \cdot I \cdot \int_0^{2\pi} dl \]
The integral \( \int_0^{2\pi} dl \) gives the circumference of the loop, which is \( 2\pi a \). Therefore, we have:
\[ F = B \cdot I \cdot (2\pi a) \]

### Step 6: Final Expression
Thus, the total force \( F \) acting on the wire is given by:
\[ F = 2\pi a I B \]

### Summary
The force on the wire due to the magnetic field from the bar magnet is:
\[ F = 2\pi a I B \]