Question

The magnetic field of a current-carrying circular loop of radius 3m at a point on the axis at a distance 4m from the center is $54\mu T$. What will be its value at the center of the loop?
A) $250\mu T$
B) $150\mu T$
C) $125\mu T$
D) $75\mu T$

Answer 1

Hint: A current traveling through a loop of wire creates a magnetic field along the axis of the loop. The direction of the field inside the loop can be found by curling the fingers of the right hand in the direction of the current through the loop; the thumb then points in the direction of the magnetic field. In the above question, we are given the magnetic field on the axis at the distance. Using the formula for the magnetic field at the center of the loop we can find the relation between two magnetic fields.

Complete step by step solution:
Express the formula for the magnetic field at the point on the axis at some distance.
$\therefore B = \dfrac{\mu _0 I r^2} {2{(x^2 + r^2)}^{3/2}}$
Where, $I$ is the current flowing in the loop,$r$ is the radius of the loop, and $x$ is the distance from the center of the loop.
Now express the formula for the magnetic field at the center of a current-carrying loop.
$\therefore B' = \dfrac{{{\mu _0}I}}{{2r}}$
Where all the symbols are in the same meaning as above.
Now take the ratio of both the magnetic field and simplify the expressions
$\therefore \dfrac{B}{{B'}} = \dfrac{ \mu _0 I r^2} {2{(x^2 + r^2)}^{3/2}} \times \dfrac{2r}{\mu _0 I }$
$ \Rightarrow \dfrac{B}{B'} = \dfrac{2 r^3 }{{(x^2 + r^2)}^{3/2}}$
Put the value 4m for $x$ , 3m for $r$ , and $54\mu T$ for B.
$\therefore \dfrac{54}{B'} = \dfrac{2 \times 3^3}{{(4^2 + 3^2)}^{3/2}}$
$ \Rightarrow \dfrac{54}{B'} = \dfrac{54}{(25)^{3/2}}$
54 will be canceled out from both sides
$ \Rightarrow \dfrac{1}{{B'}} = \dfrac{1}{{125}}$
$ \Rightarrow B' = 125\mu T$

Hence the correct option is option C.

Note: Every current-carrying wire creates some magnetic field. When we talk about a current-carrying loop the magnetic field at the center is given by $B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi {r^2}I}}{{{r^3}}}$ but we used it by simplifying. When we desire to calculate the magnetic field at a point on the axis distanced from the center $x$, then we put the distance of the point from the circumference of the loop which, by the Pythagoras theorem, is $\sqrt {{x^2} + {r^2}} $.