The mag. field due to a current carrying circular Loop of radius $3 \mathrm{~cm}\( at a point on the axis at a distance of \)4 \mathrm{~cm}$ from the centre is \(54 \mu \mathrm{T}\) what will be its value at the centre of the LOOP. (a) \(250 \mu \mathrm{T}\) (b) \(150 \mu \mathrm{T}\) (c) \(125 \mu \mathrm{T}\) (d) \(75 \mu \mathrm{T}\)

We are given the following information: - Radius of the circular loop: \( r = 3 ~ cm = 0.03 ~ m \) - Distance from the center of the loop to the point on the axis where the magnetic field is given: \( d = 4 ~ cm = 0.04 ~ m \) - Magnetic field at the given point: \( B = 54 ~ \mu T = 54 \times 10^{-6} ~ T \) The formula for the magnetic field at a point on the axis of a current-carrying circular loop is: \[ B = \dfrac{\mu_0 I r^2}{2(r^2 + x^2)^\frac{3}{2}} \] where: - \( B \) is the magnetic field at the point on the axis - \( \mu_0 \) is the permeability of free space: \( \mu_0 = 4 \pi \times 10^{-7} ~ T ~ m / A \) - \( I \) is the current in the loop - \( r \) is the radius of the loop - \( x \) is the distance from the center of the loop to the point on the axis

Using the given values and the formula for the magnetic field, we can solve for the current in the loop: \[ 54 \times 10^{-6} = \dfrac{4 \pi \times 10^{-7} I \times (0.03)^2}{2((0.03)^2 + (0.04)^2)^\frac{3}{2}} \] \[ I = \dfrac{54 \times 10^{-6} \times 2((0.03)^2 + (0.04)^2)^\frac{3}{2}}{4 \pi \times 10^{-7} \times (0.03)^2} \]

Now that we have the current, we can find the magnetic field at the center of the loop using the same formula by substituting \(x = 0\): \[ B_\text{center} = \dfrac{\mu_0 I r^2}{2(r^2 + 0^2)^\frac{3}{2}} \] Use the calculated value of \(I\) to find the magnetic field at the center of the loop.

Based on the calculated value of \( B_\text{center} \), choose the correct answer from the given options: (a) \( 250 ~ \mu T \) (b) \( 150 ~ \mu T \) (c) \( 125 ~ \mu T \) (d) \( 75 ~ \mu T \)