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 the magnetic field will be $\dfrac{1}{{2\sqrt 2 }}$ times at the centre?
A)2R \\\ B)\dfrac{{3R}}{2} \\\ C)R \\\ D)\dfrac{R}{2} \\\

Answer 1

Hint : The magnetic effect on electric currents, moving electric charges and magnetic materials is defined by a magnetic field, which is a vector field. In a magnetic field, a moving charge encounters a force that is perpendicular to both its own velocity and the magnetic field.

Complete Step By Step Answer:
The magnetic field at the axis due to the circular current is given by,
${B_{axis}} = \dfrac{{{\mu _0}}}{{4\pi }}\left( {\dfrac{{2\pi NI{R^2}}}{{{{({x^2} + {R^2})}^{\dfrac{3}{2}}}}}} \right)$
Where $N$ represents the number of turns in the coil,
$I$ represents the current flowing in the coil,
$R$ represents the radius of the circular current carrying coil, and
$x$ represents the distance from the centre of the coil on its axis
We consider $N = 1$ , and so we get,
${B_{axis}} = \dfrac{{{\mu _0}}}{{4\pi }}\left( {\dfrac{{2\pi I{R^2}}}{{{{({x^2} + {R^2})}^{\dfrac{3}{2}}}}}} \right)$
We can divide $2\pi$ by $4\pi$ , and then we get
${B_1} = \dfrac{{{\mu _0}I{R^2}}}{{2{{({R^2} + {x^2})}^{\dfrac{3}{2}}}}}$
${B_2} = \dfrac{{{\mu _0}I}}{{2R}}$
$\dfrac{1}{{2\sqrt 2 }}{B_2} = {B_1}$
Now we substitute the values of ${B_1}$ and ${B_2}$ in the above equation.
$\left( {\dfrac{1}{{2\sqrt 2 }}} \right)\left( {\dfrac{{{\mu _0}I}}{{2R}}} \right) = \dfrac{{{\mu _0}I{R^2}}}{{2{{({R^2} + {x^2})}^{\dfrac{3}{2}}}}}$
Since we have many terms same on both side, it gets cancelled out, and then we get,
$\dfrac{1}{{2\sqrt 2 {R^3}}} = \dfrac{1}{{{{\left( {{R^2} + {x^2}} \right)}^{\dfrac{3}{2}}}}}$
Now we take the reciprocal on both sides, we get,
$2\sqrt 2 {R^3} = {\left( {{R^2} + {x^2}} \right)^{\dfrac{3}{2}}}$
$2\sqrt 2$ can be written as ${2^{\dfrac{3}{2}}}$ . Also, we multiply $\dfrac{2}{3}$ on the exponent on both sides and we get,
$2{R^2} = {R^2} + {x^2} \\\ \Rightarrow 2{R^2} - {R^2} = {x^2} \\\ \Rightarrow {R^2} = {x^2} \\\$
On taking square roots on both sides, we get,
$x = R$
The distance from the centre of the coil on its axis when the intensity of magnetic field will be $\dfrac{1}{{2\sqrt 2 }}$ times at the centre is equal to $R$ .
Therefore, the correct option is $C)R$ .

Note :
Permanent magnetization exists in the Earth's crust, and the Earth's centre produces its own magnetic field, which sustains the majority of the field we calculate at the surface. As a result, the Earth can be defined as a "magnet."