Question

The radius of a circular current carrying coil is $R$. At what distance from the centre of the coil on its axis, the intensity of magnetic field will be $\frac{1}{2\sqrt{2}}$ times that at the centre?

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

Answer 1

Answer: (C)

$R$

Magnetic field at the axis due to circular current

$B_{axis}=\frac{{\mu }_0}{4\pi }\frac{2\pi Ni{r}^2}{(x^2+r^2{)}^{3/2}}$

$N\to$ no. of turn

$B_1=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$

$B_2=\frac{{\mu }_{0}I}{2R}$

$\therefore \frac{1}{2\sqrt{2}}{B}_2=B_1$

$\Rightarrow \left(\frac{1}{2\sqrt{2}}\right)\left(\frac{{\mu }_{0}i}{2R}\right)=\frac{{\mu }_{0}i{R}^2}{2(R^2+x^2{)}^{3/2}}$

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

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

$R^2=x^2$

$x=R$