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At what distance on the axis, from the centre of a circular current carrying coil of radius ${{r}}$, the magnetic field becomes $(\dfrac{1}{8})^{th}$ of the magnetic field at centre?
A) $\sqrt {{2}} {{r}}$
B) $2^{3/2} r$
C) $\sqrt{3} r$
D) $3 \sqrt{{2}} {{r}} $

seo-qna
Last updated date: 22nd Jun 2024
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Answer
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Hint: First of all, write the formula of magnetic field on the axis of the circular coil i.e. $B = \dfrac{\mu _0 i r^2} {2(r^2 + z^2)^{3/2}}$ and then write the formula of magnetic field from the centre of a circular current carrying coil i.e. ${{B' = }}\dfrac{{{{{\mu }}_{{0}}}{{i}}}}{{{{2r}}}}$. Use these two formulas and the given relation in the question and then equate.

Complete step by step solution:
Given: Radius of current carrying coil is ${{r}}$
Magnetic field at the axis is $(\dfrac{1}{8})^{th}$ times to that of the magnetic field at centre
To find: Distance on the axis
Formula for magnetic field on the axis of the circular coil is given by:
$B = \dfrac{\mu _0 i r^2} {2(r^2 + z^2)^{3/2}}$
Formula for magnetic at the center of the circular coil is given by
${{B' = }}\dfrac{{{{{\mu }}_{{0}}}{{i}}}}{{{{2r}}}}$
According to the given question, ${{B = }}\dfrac{{{1}}}{{{8}}}{{B'}}$
On substituting the values in above relation, we get
$\Rightarrow B = \dfrac{\mu _0 i r^2} {2(r^2 + z^2)^{3/2}}$ = $\dfrac{{{{{\mu }}_{{0}}}{{i}}}}{{{{2r}}}}$
On simplification, we get
$\Rightarrow 8r^3 = (r^2 + z^2)^{3/2}$
On rearranging the terms and on further simplification, we get
$\Rightarrow (4r^2)^3 = (r^2 + z^2)^3$
Taking cube root both sides, we get
$\Rightarrow 4r^2 = r^2 +z^2$
Again on rearranging terms, we get
$\Rightarrow z = \sqrt{{3}} r$
Thus, at distance, $z = \sqrt{{3}} r$ on the axis, from the centre of a circular current carrying coil of radius${{r}}$, the magnetic field becomes $(\dfrac{1}{8})^{th}$ of the magnetic field at centre.

Therefore, option (C) is the correct choice.

Note: The value of the magnetic field varies along the axis of the coil as at the centre of the coil, the magnetic field will be uniform. Just as the location of the point increases from the centre of the coil, then the value of the magnetic field decreases where ${{{\mu }}_{{0}}}$ the value of absolute permeability in free space. However, the horizontal component of the earth's magnetic field varies greatly over the surface of the earth.