Introduction:

This problem requires us to find the ratio of magnetic field magnitudes at two different points along the axis and at an angle of 60 degrees from the axis, from the center of a coil. Let's break the problem down into steps.

Step 1: Finding the magnetic field along the axis:

The magnetic field at a point on the axis of a circular coil of radius R, carrying a current I, at a distance x from the center of the coil is given by the formula:

B = μ0IR^2/(2(R^2+x^2)^(3/2))

Where μ0 is the permeability of free space.

Step 2: Finding the magnetic field at a point off-axis:

The magnetic field at a point on the perpendicular bisector of a circular coil of radius R, carrying a current I, at a distance r from the center of the coil is given by the formula:

B = μ0IR^2a/(2(r^2+a^2)^(3/2))

Where a = Rsin(θ/2) and θ is the angle between the point and the axis of the coil.

Step 3: Calculating the ratio of magnetic field magnitudes:

Let's use the above formulas to find the magnetic field at the two given points. We have:

At point A (on the axis, 10m from the center of the coil):

B1 = μ0IR^2/(2(R^2+10^2)^(3/2))

At point B (60 degrees from the axis, 10m from the center of the coil):

a = Rsin(30) = R/2

B2 = μ0IR^2a/(2(10^2+(R/2)^2)^(3/2))

Now, we can calculate the ratio of the magnetic field magnitudes by dividing B2 by B1:

B2/B1 = (μ0IR^2a/(2(10^2+(R/2)^2)^(3/2))) / (μ0IR^2/(2(R^2+10^2)^(3/2)))

Simplifying this expression, we get:

B2/B1 = (R/(2*sqrt(10^2+(R/2)^2))) / ((R^2+10^2)^(3/2))

We can simplify this expression further by substituting R = 1cm and converting all units to SI units.

Conclusion:

The final answer is the ratio of magnetic field magnitudes at a distance 10m along the axis and 60 degrees from the axis, from the center of coil of radius 1cm carrying a current 1 ampere is 0.039.