Question

A charge in the circular plates capacitor with radius of plates 5.0 cm depends on time as Q=1.0+5.00t^2-3.0t, where constants have units of C; C/s ^2 and C/s. At what distance from a center inside capacitor induced magnetic field at t= 1 s is equal to magnetic field outside 7.0 cm from the edge? Give answer in cm.

Answer 1

Answer:

Distance (d) = 2.08 cm

Explanation:

From the given information:

A diagrammatic illustration is sketched and the image is attached below:

Suppose 'i' be the total current at any instant.

Then:

$i = \dfrac{dQ}{dt}$

where Q = 1.0 +5.00t² - 3.0 t

$i = \dfrac{d}{dt}(1.0+5.00t^2-3.0t)$

i = 10.00t - 3.0

At t = 1s

i = 10.00(1) - 3.0

i = 7.00 amperes

Since the current at P is situated in the outside, the magnetic field B can be computed as:

$B_{out} = \dfrac{\mu_oi}{2 \pi r}$

$B_{out} = \dfrac{4 \pi \times 10^{-7}(7.00 \ A) }{2 \pi (12.0 \times 10^{-2} \ m)}$

$B_{out} = 1.1667 \times 10^{-5} \ T$

Taking the ampere loop with radius (d) into consideration:

The current through the loop can be expressed as:

$i' = (i)(\dfrac{\pi d^2}{\pi R^2})$

$i' = (i)(\dfrac{d^2}{ R^2})$

Thus; the magnetic field at P' i.e. the inside point is :

$B_{in} = \dfrac{\mu_o i'}{2 \pi d}$

$B_{in} = \dfrac{\mu_o }{2 \pi d}(i \dfrac{d^2}{R^2})$

$B_{in} = \dfrac{\mu_oi }{2 \pi d}(\dfrac{d^2}{R^2})$

$B_{in} = \dfrac{\mu_oi }{2 \pi }(\dfrac{d}{R^2})$

$B_{in} = \dfrac{4 \pi \times 10^{-7} \times 7.00 }{2 \pi }(\dfrac{d}{(5.0 \times 10^{-2})^2})$

$B_{in} = 1.4 \times 10^{-6} \times (\dfrac{d}{0.0025})$

$B_{in} = 5.6 \times 10^{-4} \ d$

From the question; Since $B_{out} = B_{in}$

Then:

$1.1667 \times 10^{-5}= 5.6 \times 10^{-4} \ d$

$d = \dfrac{1.1667 \times 10^{-5}}{5.6 \times 10^{-4}}$

d = 0.0208

d = 2.08 cm