For a circular coil of radius a , at what distance x along its axis the magnetic field is exactly 27 1 of its value at the center?

Question

For a circular coil of radius a , at what distance x along its axis the magnetic field is exactly 27 1 ​ of its value at the center?

TestHub · Accepted Answer

The magnetic field at the center of a circular coil of radius $a$ is given by $B_{c} = \frac{μ _{0} I}{2 a}$ .

The magnetic field at a distance $x$ along the axis is given by $B_{x} = \frac{μ _{0} I a ^{2}}{2 ( a ^{2} + x ^{2} ) ^{3/2}}$ .

Given that $B_{x} = \frac{1}{27} B_{c}$ , we have:

$\frac{μ _{0} I a ^{2}}{2 ( a ^{2} + x ^{2} ) ^{3/2}} = \frac{1}{27} (\frac{μ _{0} I}{2 a})$

$\frac{a ^{3}}{( a ^{2} + x ^{2} ) ^{3/2}} = \frac{1}{27}$

Taking the cube root on both sides:

$\frac{a}{a ^{2} + x ^{2}} = \frac{1}{3}$

Squaring both sides:

$\frac{a ^{2}}{a ^{2} + x ^{2}} = \frac{1}{9}$

$9 a^{2} = a^{2} + x^{2}$

$x^{2} = 8 a^{2}$

$x = 22 a$

Answer: $22 a$

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