Chapter 22, Problem 42QAP

To determine

The electric field from time intervals 0 to 4 ms, 4 to 10 ms and 10 to 14 ms.

Answer to Problem 42QAP

The mass of the mediating particle. is, $E_{0\to 4}=-6.62\times {10}^{-4}\text{N/C}$, $E_{4\to 10}=4.4\times {10}^{-6}\text{N/C}$ and $E_{10\to 14}=0\text{N/C}$

Explanation of Solution

Given:

Diameter of ring, $d=1.5\text{cm}$

Radius of the ring, $r=0.75\text{cm=0}\text{.75}\times {\text{10}}^{-2}\text{m}$

Formula used:

Electric field is given by,

  $E=\frac{\varphi }{\Delta t}$

Where,

  $E$ =Electric field

  $\varphi$ = Electric flux

  $\Delta t$ =Time interval

  $R$ =Radius of earth

Electric field form t=0 to t=4ms is given by,

  $\begin{array}{l}{E}_{0\to 4}=\frac{\Delta \varphi }{\Delta t}\\ \text{But, }\Delta \varphi \text{=}{\left(BA\right)}_{t=0}-{\left(BA\right)}_{t=4}=0-{\left(BA\right)}_{t=4}\\ \therefore E_{0\to 4}=-\frac{{\left(BA\right)}_{t=4}}{\Delta t}\\ E_{0\to 4}=-\frac{B\left(\pi r^2\right)}{\Delta t}\\ E_{0\to 4}=-\frac{15\times {10}^{-3}\left(3.14\times {\left(\text{0}\text{.75}\times {\text{10}}^{-2}\right)}^2\right)}{4\times {\text{10}}^{-3}}\\ E_{0\to 4}=-6.62\times {10}^{-4}\text{N/C}\end{array}$

Electric field form t=4ms to t=10ms is given by,

  $\begin{array}{l}{E}_{4\to 10}=\frac{\Delta \varphi }{\Delta t}\\ \text{But, }\Delta \varphi \text{=}{\left(BA\right)}_{t=4}-{\left(BA\right)}_{t=10}\\ \therefore E_{4\to 10}=\frac{{\left(BA\right)}_{t=4}-{\left(BA\right)}_{t=10}}{\Delta t}\\ E_{4\to 10}=\frac{\left(\pi r^2\right)\left(B_{t=4}-B_{t=10}\right)}{\Delta t}\\ E_{4\to 10}=\frac{\left(3.14\times {\left(\text{0}\text{.75}\times {\text{10}}^{-2}\right)}^2\right)\left(15\times {10}^{-3}-\left(-10\times {10}^{-3}\right)\right)}{6\times \times {10}^{-3}}\\ E_{4\to 10}=4.4\times {10}^{-6}\text{N/C}\end{array}$

Electric field form t=10ms to t=14ms is given by,

  $\begin{array}{l}{E}_{10\to 14}=\frac{\Delta \varphi }{\Delta t}\\ \text{But, }\Delta \varphi \text{=}{\left(BA\right)}_{t=10}-{\left(BA\right)}_{t=14}\\ \therefore E_{10\to 14}=\frac{{\left(BA\right)}_{t=10}-{\left(BA\right)}_{t=14}}{\Delta t}\\ E_{10\to 14}=\frac{\left(\pi r^2\right)\left(B_{t=10}-B_{t=14}\right)}{\Delta t}\\ E_{10\to 14}=\frac{\left(3.14\times {\left(\text{0}\text{.75}\times {\text{10}}^{-2}\right)}^2\right)\left(-10\times {10}^{-3}-\left(-10\times {10}^{-3}\right)\right)}{6\times \times {10}^{-3}}\\ E_{10\to 14}=0\text{N/C}\end{array}$