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Question: A circular current-carrying coil has a radius R. The distance from the center of the coil, on the ax...

A circular current-carrying coil has a radius R. The distance from the center of the coil, on the axis, where B will be 18\dfrac{1}{8} of its value at the center of the coil is:
A) R3\dfrac{R}{{\sqrt 3 }}
B) 3R\sqrt 3 R
C) 23R2\sqrt 3 R
D) 2R3\dfrac{{2R}}{{\sqrt 3 }}

Explanation

Solution

We know that a circular current-carrying coil will have a magnetic field along its axis. For any circular current carrying current, the magnetic field along its axis will be given by B=μ0IR22(R2+a2)32B = \dfrac{{{\mu _0}I{R^2}}}{{2{{\left( {{R^2} + {a^2}} \right)}^{\dfrac{3}{2}}}}} . In the given question, first, we need to find the magnetic field at the center which can be given by simply putting the value of a=0a = 0 . Now, we can calculate the distance aa by equating the condition given in the question mathematically.

Complete step by solution:
According to the question, we are given:
RR = radius of circular current-carrying coil
Now, the magnetic field along the axis of the circular coil is given by:
B=μ0IR22(R2+a2)32B = \dfrac{{{\mu _0}I{R^2}}}{{2{{\left( {{R^2} + {a^2}} \right)}^{\dfrac{3}{2}}}}}
Where,
μ0{\mu _0} is the permeability of free space
II is the current flowing in the coil
RR is the radius of circular current-carrying coil
aa is the distance from the center of the coil on the axis where B is calculated
Let this be equation 1.
We need to find the magnetic field at the center of the circular coil i.e B0{B_0} .
Substituting the a=0a = 0 in equation 1, we get
B0=μ0I2R{B_0} = \dfrac{{{\mu _0}I}}{{2R}}
According to the given data in the question, B=18B0B = \dfrac{1}{8}{B_0}
Now, putting values of BB and B0{B_0}, we get

μ0IR22(R2+a2)32=18μ0I2R 8R3=(R2+a2)32 64R6=(R2+a2)3 (4R2)3=(R2+a2)3 4R2=R2+a2 a=3R \Rightarrow \dfrac{{{\mu _0}I{R^2}}}{{2{{\left( {{R^2} + {a^2}} \right)}^{\dfrac{3}{2}}}}} = \dfrac{1}{8}\dfrac{{{\mu _0}I}}{{2R}} \\\ \Rightarrow 8{R^3} = {\left( {{R^2} + {a^2}} \right)^{\dfrac{3}{2}}} \\\ \Rightarrow 64{R^6} = {\left( {{R^2} + {a^2}} \right)^3} \\\ \Rightarrow {\left( {4{R^2}} \right)^3} = {\left( {{R^2} + {a^2}} \right)^3} \\\ \Rightarrow 4{R^2} = {R^2} + {a^2} \\\ \Rightarrow a = \sqrt 3 R \\\

Therefore, the distance from the center of the coil, on the axis, where B will be 18\dfrac{1}{8} of its value at the center of the coil is 3R\sqrt 3 R .

Hence, option (B) is correct.

Note: These types of questions can be solved by using the direct formula for the magnetic field on a circular current-carrying coil. We should always be cautious doing such calculations because a small calculation mistake can lead us to an incorrect answer. Also, we need to be precise with the details given in the question.