Question

A circular coil carrying current 'I' has a radius 'R' and magnetic field at the center is 'B'. At what distance from the center along the axis of the same coil, the magnetic field will be $\frac{B}{8}$?

• A. $R\sqrt{2}$
• B. $R\sqrt{3}$
• C. $2R$
• D. $3R$

Answer 1

Answer: (B)

$R\sqrt{3}$

We know that at the axis of a current carrying coil at distance $x$ from the center , magnetic field $B$ is given by-

$B_1=\frac{{\mu }_{o}I{R}^2}{2(R^2+x^2{)}^3/2}$

And $B=\frac{{\mu }_{o}I}{2R}$

$B_1=\frac{B}{8}$

$\frac{{\mu }_{o}I{R}^2}{2(R^2+x^2{)}^3/2}=\frac{{\mu }_{o}I}{16R}$

$⟹8R^3=(R^2+x^2{)}^{3/2}$

$⟹2R=(R^2+x^2{)}^{1/2}$

$⟹4R^2=R^2+x^2$

$⟹x^2=3R^2$

$⟹x=R\sqrt{3}$

Hence, answer is option-(B).