Explanation

(a) The magnetic field at a point on the axis of a circular loop at a distance $x$ from the centre is
$B = \frac{\mu_{0}iR^{2}}{2\left( R^{2} + x^{2} \right)^{3/2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Given, $B = 54\ \mu\ T,\ x = 4\ cm,\ R = 3\ cm$
Putting the given values in Eq. (i), we get
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ 54 = \frac{\mu_{0}i \times (3)^{2}}{2\left( 3^{2} + 4^{2} \right)^{3/2}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ 54 = \frac{9\mu_{0}i}{2(25)^{3/2}} = \frac{9\mu_{0}i}{2 \times (5)^{3}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \mu_{0}i = \frac{54 \times 2 \times 125}{9}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mu_{0}i = 1500\ \mu\ T - cm\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Now, putting $x = 0$ in Eq. (i), magnetic field at the centre of loop is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B = \frac{\mu_{0}iR^{2}}{2R^{3}} = \frac{\mu_{0}i}{2R} = \frac{1500}{2 \times 3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 250\ \mu T$ [From Eq. (ii)]