Answer:
[tex]$ 1.81 \times 10^{-7} \ T$[/tex]
Explanation:
Given :
r = 1.50 m
R = 1 m
[tex]$\frac{dE}{dt}$[/tex] = [tex]$ 4.88 \times 10^{10} $[/tex] V / m s
Therefore the displacement current is
[tex]$ I_d = \epsilon_0. \frac{dE}{dt} . A $[/tex]
= [tex]$ \epsilon_0. \frac{dE}{dt} . \pi R^2 $[/tex]
Now according to law
[tex]$ B= \frac{\mu_0I_d.\frac{dE}{dt}.\pi R^2}{2 \pi r}$[/tex]
= [tex]$ \frac{\mu_0I_d.\frac{dE}{dt}. R^2}{2 r}$[/tex]
= [tex]$ \frac{4 \pi \times 10^{-7} \times 8.85 \times 10^{-12} \times 4.88 \times 10^{10} \times 1^2}{2 \times 1.5} $[/tex]
= [tex]$ 1.81 \times 10^{-7} \ T$[/tex]
Therefore, the magnetic field strength at 1.50 m from the center of the ring is [tex]$ 1.81 \times 10^{-7} \ T$[/tex].
