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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
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B
3R2
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C
R
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D
R2
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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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