
The magnetic field at two points on the axis of a circular coil at a distance of 0.05m and 0.2m from the center are in the ratio 2:1. The radius of the coil is:
a)1.0m
b)0.24m
c)0.15m
d)0.3m
Answer
560.1k+ views
Hint: In the above question it is asked to determine the radius of a particular coil. We are given that the magnetic field at two points on the axis of a circular coil at a distance of 0.05m and 0.2m from the center are in the ratio 2:1. Hence we will use the expression for magnetic field at a point on the axis of the coil considering the above two cases and then accordingly take the ratio in order to determine the radius of the coil.
Formula used: $B=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{x}^{2}}+{{R}^{2}})}^{3/2}}}$
Complete step by step answer:
Let us say we have a coil of radius R and the current through the coil is ‘i’. Let us say we wish to calculate the field at a point along the axis of the coil passing through its center. The magnetic field (B) at a distance ‘x’ from the coil is given by,
$B=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{x}^{2}}+{{R}^{2}})}^{3/2}}}$
Let us first determine the magnetic field (${{B}_{1}}$) at point 0.05m from the coil.
$\begin{align}
& {{B}_{1}}=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{0.05}^{2}}+{{R}^{2}})}^{3/2}}} \\
& {{B}_{1}}=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{(0.0025+{{R}^{2}})}^{3/2}}}...(1) \\
\end{align}$
Now let us determine the magnetic field (${{B}_{2}}$) at point 0.2m from the center of the coil,
$\begin{align}
& {{B}_{2}}=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{0.2}^{2}}+{{R}^{2}})}^{3/2}}} \\
& {{B}_{2}}=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{(0.04+{{R}^{2}})}^{3/2}}}...(2) \\
\end{align}$
Dividing equation 1 by 2 we get,
$\begin{align}
& \dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{(0.0025+{{R}^{2}})}^{3/2}}}}{\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{(0.04+{{R}^{2}})}^{3/2}}}}\text{, }\because \dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{2}{1} \\
& \Rightarrow \dfrac{2}{1}=\dfrac{{{(0.04+{{R}^{2}})}^{3/2}}}{{{(0.0025+{{R}^{2}})}^{3/2}}} \\
& \Rightarrow {{\left( \dfrac{4}{1} \right)}^{1/3}}=\dfrac{(0.04+{{R}^{2}})}{(0.0025+{{R}^{2}})} \\
& \Rightarrow 1.58(0.0025+{{R}^{2}})=(0.04+{{R}^{2}}) \\
& \Rightarrow 0.0039+1.58{{R}^{2}}=0.04+{{R}^{2}} \\
& \Rightarrow 0.58{{R}^{2}}=0.04-0.0039=0.0361 \\
& \Rightarrow R=\sqrt{\dfrac{0.0361}{0.58}}=0.24m \\
\end{align}$
So, the correct answer is “Option b”.
Note: ${{\mu }_{\circ }}$ in the above expression represents the permeability of free space i.e. in vacuum. This constant depends on the medium where the magnetic field is present. If we consider the expression for magnetic field due to a coil, we can imply that the field is maximum at the center where x goes to zero. Similarly, we can say that at infinity the field due to the coil will be zero.
Formula used: $B=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{x}^{2}}+{{R}^{2}})}^{3/2}}}$
Complete step by step answer:
Let us say we have a coil of radius R and the current through the coil is ‘i’. Let us say we wish to calculate the field at a point along the axis of the coil passing through its center. The magnetic field (B) at a distance ‘x’ from the coil is given by,
$B=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{x}^{2}}+{{R}^{2}})}^{3/2}}}$
Let us first determine the magnetic field (${{B}_{1}}$) at point 0.05m from the coil.
$\begin{align}
& {{B}_{1}}=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{0.05}^{2}}+{{R}^{2}})}^{3/2}}} \\
& {{B}_{1}}=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{(0.0025+{{R}^{2}})}^{3/2}}}...(1) \\
\end{align}$
Now let us determine the magnetic field (${{B}_{2}}$) at point 0.2m from the center of the coil,
$\begin{align}
& {{B}_{2}}=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{0.2}^{2}}+{{R}^{2}})}^{3/2}}} \\
& {{B}_{2}}=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{(0.04+{{R}^{2}})}^{3/2}}}...(2) \\
\end{align}$
Dividing equation 1 by 2 we get,
$\begin{align}
& \dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{(0.0025+{{R}^{2}})}^{3/2}}}}{\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{(0.04+{{R}^{2}})}^{3/2}}}}\text{, }\because \dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{2}{1} \\
& \Rightarrow \dfrac{2}{1}=\dfrac{{{(0.04+{{R}^{2}})}^{3/2}}}{{{(0.0025+{{R}^{2}})}^{3/2}}} \\
& \Rightarrow {{\left( \dfrac{4}{1} \right)}^{1/3}}=\dfrac{(0.04+{{R}^{2}})}{(0.0025+{{R}^{2}})} \\
& \Rightarrow 1.58(0.0025+{{R}^{2}})=(0.04+{{R}^{2}}) \\
& \Rightarrow 0.0039+1.58{{R}^{2}}=0.04+{{R}^{2}} \\
& \Rightarrow 0.58{{R}^{2}}=0.04-0.0039=0.0361 \\
& \Rightarrow R=\sqrt{\dfrac{0.0361}{0.58}}=0.24m \\
\end{align}$
So, the correct answer is “Option b”.
Note: ${{\mu }_{\circ }}$ in the above expression represents the permeability of free space i.e. in vacuum. This constant depends on the medium where the magnetic field is present. If we consider the expression for magnetic field due to a coil, we can imply that the field is maximum at the center where x goes to zero. Similarly, we can say that at infinity the field due to the coil will be zero.
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