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

Answer:

$1.81 \times 10^{-7} \ T$

Explanation:

Given :

r = 1.50 m

R = 1  m

$\frac{dE}{dt}$ = $4.88 \times 10^{10}$  V / m s

Therefore the displacement current is

$I_d = \epsilon_0. \frac{dE}{dt} . A$

    = $\epsilon_0. \frac{dE}{dt} . \pi R^2$

Now according to law

$B= \frac{\mu_0I_d.\frac{dE}{dt}.\pi R^2}{2 \pi r}$

  = $\frac{\mu_0I_d.\frac{dE}{dt}. R^2}{2 r}$

  = $\frac{4 \pi \times 10^{-7} \times 8.85 \times 10^{-12} \times 4.88 \times 10^{10} \times 1^2}{2 \times 1.5}$

  = $1.81 \times 10^{-7} \ T$

Therefore, the magnetic field strength at 1.50 m from the center of the ring is $1.81 \times 10^{-7} \ T$.