Question

The radius of a circular current carrying coil on its. At what distance from the centre of the coil on its axis the magnetic field will be $\frac{1}{3\sqrt{3}}$ times to that at centre?

Answer 1

R\sqrt{2}

Answer 2

The magnetic field at the center of a circular current-carrying coil of radius $R$ is given by:

$B_c = \frac{\mu_0 I}{2R}$

The magnetic field at a distance $x$ from the center of the coil on its axis is given by:

$B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$

According to the problem, the magnetic field at distance $x$ is $\frac{1}{3\sqrt{3}}$ times the magnetic field at the center:

$B_x = \frac{1}{3\sqrt{3}} B_c$

Substitute the expressions for $B_x$ and $B_c$:

$\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{3\sqrt{3}} \left( \frac{\mu_0 I}{2R} \right)$

Cancel out the common terms $\frac{\mu_0 I}{2}$ from both sides:

$\frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{3\sqrt{3} R}$

Rearrange the equation to isolate the term with $x$:

$3\sqrt{3} R \cdot R^2 = (R^2 + x^2)^{3/2}$

$3\sqrt{3} R^3 = (R^2 + x^2)^{3/2}$

We know that $3\sqrt{3}$ can be written as $(\sqrt{3})^3$. So, the left side becomes:

$(\sqrt{3})^3 R^3 = (R^2 + x^2)^{3/2}$

$(\sqrt{3} R)^3 = (R^2 + x^2)^{3/2}$

To simplify, take the cube root of both sides:

$(\sqrt{3} R) = (R^2 + x^2)^{1/2}$

Now, square both sides to eliminate the square root:

$(\sqrt{3} R)^2 = (R^2 + x^2)$

$3R^2 = R^2 + x^2$

Solve for $x^2$:

$x^2 = 3R^2 - R^2$

$x^2 = 2R^2$

Finally, solve for $x$:

$x = \sqrt{2R^2}$

$x = R\sqrt{2}$

The distance from the center of the coil on its axis where the magnetic field is $\frac{1}{3\sqrt{3}}$ times that at the center is $R\sqrt{2}$.