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

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 2 2 a

The magnetic field at the center of a circular coil of radius

a

is given by

B_c = \dfrac{\mu_0 I}{2a}

.

The magnetic field at a distance

x

along the axis is given by

B_x = \dfrac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}

.

Given that

B_x = \dfrac{1}{27} B_c

, we have:

\dfrac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}} = \dfrac{1}{27} \left( \dfrac{\mu_0 I}{2a} \right)<br><br>\dfrac{a^3}{(a^2 + x^2)^{3/2}} = \dfrac{1}{27}<br><br>Taking the cube root on both sides:<br><br>\dfrac{a}{\sqrt{a^2 + x^2}} = \dfrac{1}{3}<br><br>Squaring both sides:<br><br>\dfrac{a^2}{a^2 + x^2} = \dfrac{1}{9}<br><br>9a^2 = a² + x²

x^2 = 8a^2

x = 2\sqrt{2}

a

<br><br>Answer: 2\sqrt{2}

a

For a circular coil of radius a , at what distance x along its axis...