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 $$\dfrac { 1 }{ 2\sqrt { 2 } }$$ times that at the centre?

A
2R
B
3R2
C
R
D
R2

Solution

The correct option is A $$R$$Magnetic field at the axis due to circular current$$B_{axis}=\dfrac{\mu_0}{4\pi}\dfrac{2\pi Nir^2}{(x^2+r^2)^{3/2}}$$$$N\rightarrow$$ no. of turn$$B_1=\dfrac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}$$$$B_2=\dfrac{\mu_0I}{2R}$$$$\therefore\dfrac{1}{2 \sqrt2} B_2= B_1$$$$\Rightarrow \left(\dfrac{1}{2\sqrt 2}\right)\left(\dfrac{\mu_0i}{2R}\right)=\dfrac{\mu_0iR^2}{2(R^2+x^2)^{3/2}}$$$$2\sqrt 2R^3=(R^2+x^2)^{3/2}$$$$2R^2=R^2+x^2$$$$R^2=x^2$$$$\boxed{x=R}$$Physics

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