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."