Answer:
[tex]B=1.21*10^{-10}T[/tex]
Explanation:
The magnitude of the magnetic field on the axis of the ring is given by:
[tex]B=\frac{\mu_0 IR^2}{2(r^2+R^2)^{\frac{3}{2}}}(1)[/tex]
[tex]\mu_0[/tex] is the permeability of free space, [tex]I[/tex] is the flowing current through the ring, [tex]R[/tex] is the ring's radius and [tex]r[/tex] is the distance to the center of the ring.
The flowing current through the ring is defined as the ring's charge divided into the time taken by the charge to complete one revolution, that is, the period [tex]T=\frac{2\pi}{\omega}[/tex]. So, we have:
[tex]I=\frac{q}{T}\\I=\frac{q}{\frac{2\pi}{\omega}}\\\\I=\frac{\omega q}{2\pi}\\I=\frac{18\frac{rad}{s}(10*10^{-6}C)}{2\pi}\\I=2.87*10^{-5}A[/tex]
Now, replacing in (1):
[tex]B=\frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(2.87*10^{-5}A)(0.115m)^2}{2((0.05m)^2+(0.115m)^2)^{\frac{3}{2}}}\\B=1.21*10^{-10}T[/tex]
