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Step-by-Step Solution
Step 1: Write the expression for the magnetic field on the axis of a circular coil
The magnetic field $B$ at a point on the axis of a circular coil of radius $R$, carrying current $i$, placed at a distance $x$ from its center is given by:
$B = \frac{\mu_{0} N i R^{2}}{2 \bigl(R^{2} + x^{2}\bigr)^{\frac{3}{2}}}$
where $N$ is the number of turns and $\mu_{0}$ is the permeability of free space.
Step 2: Express the fields at the two given points
Two points on the axis are at distances $x_{1} = 0.05 \text{ m}$ and $x_{2} = 0.2 \text{ m}$ from the center. Hence, the magnetic fields at these two points are:
$B_{1} = \frac{\mu_{0} N i R^{2}}{2\,[R^{2} + (0.05)^{2}]^{\frac{3}{2}}}$
$B_{2} = \frac{\mu_{0} N i R^{2}}{2\,[R^{2} + (0.2)^{2}]^{\frac{3}{2}}}$
Step 3: Form the ratio of the two magnetic fields
We are given that $B_{1} : B_{2} = 8 : 1$. So,
$\frac{B_{1}}{B_{2}} = \frac{\frac{\mu_{0} N i R^{2}}{2\,[R^{2} + (0.05)^{2}]^{\frac{3}{2}}}}{\frac{\mu_{0} N i R^{2}}{2\,[R^{2} + (0.2)^{2}]^{\frac{3}{2}}}}
= \frac{[R^{2} + (0.2)^{2}]^{\frac{3}{2}}}{[R^{2} + (0.05)^{2}]^{\frac{3}{2}}} = 8.$
Step 4: Simplify and solve for $R$
Raising both sides to the power of $2/3$ to eliminate the exponent $3/2$, we get:
$(8)^{\frac{2}{3}}
= \frac{R^{2} + (0.2)^{2}}{R^{2} + (0.05)^{2}}.
$
Note that $(8)^{\frac{2}{3}} = 4$. Hence,
$4 \bigl(R^{2} + 0.05^{2}\bigr) = R^{2} + 0.2^{2}.
$
Substitute $0.05^{2} = 0.0025$ and $0.2^{2} = 0.04$:
$4(R^{2} + 0.0025) = R^{2} + 0.04.
$
Step 5: Rearrange and find the radius $R$
$4R^{2} + 4 \times 0.0025 = R^{2} + 0.04
$
$4R^{2} + 0.01 = R^{2} + 0.04
$
$3R^{2} = 0.04 - 0.01 = 0.03
$
$R^{2} = \frac{0.03}{3} = 0.01
$
$R = \sqrt{0.01} = 0.1\ \text{m}.
$
Final Answer
The radius of the coil is $0.1\ \text{m}$.
