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Step-by-Step Solution
Step 1: Recall the magnetic field on the axis of a circular loop
The magnetic field at a point on the axis of a current-carrying circular loop of radius $a$, at a distance $x$ from its center, is given by:
$B = \frac{\mu_0\,i\,a^2}{2\left(x^2 + a^2\right)^{\frac{3}{2}}}$
Step 2: Recall the magnetic field at the center of the loop
The magnetic field at the center of the loop (i.e., $x=0$) is given by:
$B' = \frac{\mu_0\,i}{2\,a}
Step 3: Relate the field on the axis to the field at the center
From the given relations, if $B$ is the field at distance $x$, and $B'$ is the field at the center, then:
$B' = B \times \frac{\left(x^2 + a^2\right)^{\frac{3}{2}}}{a^3}
This formula allows us to find $B'$ if $B$ at some known point on the axis (with distance $x$ from the center) is given.
Step 4: Substitute the given values
We are given:
Radius of the loop $a = 3\text{ cm}$
Distance from center on the axis $x = 4\text{ cm}$
Magnetic field at this point $B = 54\,\mu T$
Substitute these into the relation for $B'$:
$B' = 54\,\mu T \times \frac{\left(4^2 + 3^2\right)^{\frac{3}{2}}}{3^3}
Here, $4^2 + 3^2 = 16 + 9 = 25$, so $\left(25\right)^{\frac{3}{2}} = 25^{1.5} = 125.$
$B' = 54\,\mu T \times \frac{125}{3 \times 3 \times 3}
= 54\,\mu T \times \frac{125}{27}
= 54 \times \frac{125}{27}\,\mu T
= 54 \times \frac{125}{27}\,\mu T
= 250\,\mu T
Step 5: Conclusion
Therefore, the magnetic field at the center of the loop is $250\,\mu T.$
