The ratio 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 that at the centre?
A. $2R$
B. $\dfrac{{3R}}{2}$
C. $R$
D. $\dfrac{R}{2}$
Answer
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Hint: In order to answer this question, first we will apply the formula of the magnetic field in terms of number of turns, distance and radius of the coil. And then we will write the expression of two consecutive terms and compare it to get the distance value.
Complete step by step answer:
Magnetic field at the axis due to circular current:
We will apply the formula of the magnetic field:-
${B_{axis}} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi Ni{r^2}}}{{{{({x^2} + {r^2})}^{\dfrac{3}{2}}}}}$
where, \[N\] is the number of turns, $x$ is the distance from the centre of the coil on its axis and ${\mu _0}$ is the permeability constant.
${B_1} = \dfrac{{{\mu _0}I{R^2}}}{{2{{({R^2} + {x^2})}^{\dfrac{3}{2}}}}}$
$\Rightarrow {B_2} = \dfrac{{{\mu _0}I}}{{2R}}$
By comparing ${B_1}$ and ${B_2}$ , we get-
$\dfrac{1}{{2\sqrt 2 }}{B_2} = {B_1}$
$ \Rightarrow (\dfrac{1}{{2\sqrt 2 }})(\dfrac{{{\mu _0}i}}{{2R}}) = \dfrac{{{\mu _0}i{R^2}}}{{2{{({R^2} + {x^2})}^{\dfrac{3}{2}}}}}$
$ \Rightarrow 2\sqrt 2 {R^3} = {({R^2} + {x^2})^{\dfrac{3}{2}}}$
$\Rightarrow 2{R^2} = {R^2} + {x^2} \\
\Rightarrow {R^2} = {x^2} \\
\therefore x = R $
Therefore, $R$ is the distance from the centre of the coil on its axis, the intensity of the magnetic field will be $\dfrac{1}{{2\sqrt 2 }}$ times that at the centre.
Hence, the correct option is C.
Note:When the current is going in the opposite direction, the magnetic field produced is generally outwards, and the coil's face acts as the north pole. The magnetic field at the arc's centre for the whole circle, the magnetic field at the centre of an arc is proportional to the angle of arc of $2\pi $ , i.e.. ${B_{arc}} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{I}{R}(\theta )$, where $I$ is the electric current carrying by the coil and $R$ is the resistance.
Complete step by step answer:
Magnetic field at the axis due to circular current:
We will apply the formula of the magnetic field:-
${B_{axis}} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi Ni{r^2}}}{{{{({x^2} + {r^2})}^{\dfrac{3}{2}}}}}$
where, \[N\] is the number of turns, $x$ is the distance from the centre of the coil on its axis and ${\mu _0}$ is the permeability constant.
${B_1} = \dfrac{{{\mu _0}I{R^2}}}{{2{{({R^2} + {x^2})}^{\dfrac{3}{2}}}}}$
$\Rightarrow {B_2} = \dfrac{{{\mu _0}I}}{{2R}}$
By comparing ${B_1}$ and ${B_2}$ , we get-
$\dfrac{1}{{2\sqrt 2 }}{B_2} = {B_1}$
$ \Rightarrow (\dfrac{1}{{2\sqrt 2 }})(\dfrac{{{\mu _0}i}}{{2R}}) = \dfrac{{{\mu _0}i{R^2}}}{{2{{({R^2} + {x^2})}^{\dfrac{3}{2}}}}}$
$ \Rightarrow 2\sqrt 2 {R^3} = {({R^2} + {x^2})^{\dfrac{3}{2}}}$
$\Rightarrow 2{R^2} = {R^2} + {x^2} \\
\Rightarrow {R^2} = {x^2} \\
\therefore x = R $
Therefore, $R$ is the distance from the centre of the coil on its axis, the intensity of the magnetic field will be $\dfrac{1}{{2\sqrt 2 }}$ times that at the centre.
Hence, the correct option is C.
Note:When the current is going in the opposite direction, the magnetic field produced is generally outwards, and the coil's face acts as the north pole. The magnetic field at the arc's centre for the whole circle, the magnetic field at the centre of an arc is proportional to the angle of arc of $2\pi $ , i.e.. ${B_{arc}} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{I}{R}(\theta )$, where $I$ is the electric current carrying by the coil and $R$ is the resistance.
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