Question

Find the magnetic field strength at 1.50 m from the center of the circular region (e.g., outside the electric-field region).

Answer 1

`$ 1.81 * 10^(-7) \ T$`

Step-by-step explanation:

Given :

r = 1.50 m

R = 1 m

`$(dE)/(dt)$` =
`$ 4.88 * 10^(10) $` V / m s

Therefore the displacement current is

`$ I_d = \epsilon_0. (dE)/(dt) . A $`

=
`$ \epsilon_0. (dE)/(dt) . \pi R^2 $`

Now according to law

`$ B= (\mu_0I_d.(dE)/(dt).\pi R^2)/(2 \pi r)$`

=
`$ (\mu_0I_d.(dE)/(dt). R^2)/(2 r)$`

=
`$ (4 \pi * 10^(-7) * 8.85 * 10^(-12) * 4.88 * 10^(10) * 1^2)/(2 * 1.5) $`

=
`$ 1.81 * 10^(-7) \ T$`

Therefore, the magnetic field strength at 1.50 m from the center of the ring is
`$ 1.81 * 10^(-7) \ T$`.