Magnetising Force in a Square Wire

Given:
Length of wire = 2.5m
Current flowing through wire = 100A

To find:
Magnetising force at the centre of the square

Solution:
Let's assume that the wire is bent to form a square of side ‘a’.
Therefore, the length of the wire will be equal to the perimeter of the square.
Given, the length of the wire = 2.5m
Perimeter of the square = 4a

Therefore, 4a = 2.5m
a = 0.625m

Now, let's assume that the current flowing through the wire is distributed uniformly throughout the wire.

The magnetic field produced by a current-carrying wire is given by the formula:
B = μI/2πr
where,
B = Magnetic field strength
μ = Permeability of free space (4π × 10^-7 H/m)
I = Current flowing through the wire
r = Distance from the wire

Let's consider a small element of length ‘dl’ on one of the sides of the square. The magnetic field produced by this element at the centre of the square is given by:
dH = μI/4πr^2 * cos(45°)
where,
dH = Magnetising force produced by the element
r = Distance between the element and the centre of the square

Since the wire is bent to form a square, the magnetising forces produced by the elements on opposite sides of the square will cancel out each other. Therefore, we need to consider only the magnetising forces produced by the elements on adjacent sides of the square.

The total magnetising force produced by two adjacent sides of the square is given by:
dHtotal = 2 * dH
dHtotal = 2 * μI/4πr^2 * cos(45°)
dHtotal = μI/2√2πr^2

The distance between the centre of the square and any one of the adjacent sides is given by:
r = a/√2

Therefore,
dHtotal = μI/2πa^2

Now, we need to find the total magnetising force produced by all four sides of the square. Since the current is distributed uniformly throughout the wire, the total magnetising force produced by all four sides will be equal to four times the magnetising force produced by two adjacent sides.

Therefore,
H = 4 * dHtotal
H = 4 * μI/2πa^2
H = μI/πa^2
H = μI/π(0.625)^2
H = 143.68 AT/m

Therefore, the magnetising force at the centre of the square is 144 AT/m (approx).

Hence, option (B) is the correct answer.