Chapter 21, Problem 18P

To determine

The charge during the operation.

Answer to Problem 18P

Solution:

$\text{5}\text{.23 C}$ charge moves past a point in the coil during operation.

Explanation of Solution

Given info:

Diameter of loop=$\text{13}\text{.2 cm}$

Radius of loop = $\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

=$\text{6}\text{.6 cm}$

= $\text{0}\text{.066 m}$

Magnetic field $\left(B\right)=\text{0}\text{.670 T}$

Diameter wire =$\text{2}\text{.25 mm}$

Radius wire= $\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

=$\text{1}\text{.125}\times {\text{10}}^{-3}$

Formulas to be used:

By combining the equations

The charge with respect to elapsed time

$Q=It$ …… (1)

By ohm’s law, the induced current

$I=\frac{E}{R}$…… (2)

The resistance is given by,

$R=\frac{\rho l}{A_w}$[A_w= Area of wire]...... (3)

By Faraday’s law, the induced emf is given by,

$\left|E\right|=\frac{A_{\varphi }\left|\Delta B\right|}{\Delta t}$ [$A_{\varphi }$=Area of loop]……(4)

Combining the equations (2), (3)and (4)

$I=\frac{\frac{A_{\varphi }\left|\Delta B\right|}{\Delta t}}{\frac{\rho l}{A_w}}$

$I=\frac{A_{\varphi }{A}_w\left|\Delta B\right|}{\rho l\Delta t}$ ……(5)

Calculation:

Substituting I value from equation in equation (1)

$Q=I\Delta t$

$=\frac{A_{\varphi }{A}_w\left|\Delta B\right|}{\rho l}$

$=\frac{\left(\pi r_{\varphi }{}^2\right)\left(\pi r_w{}^2\right)\left|\Delta B\right|}{\rho 2\pi r_{\varphi }}$$[l=2\pi rA=\pi r^2]$

Q $=\frac{\left(r_{\varphi }\right)\pi r_w{}^2\left|\Delta B\right|}{2\rho }$

Q $=\frac{\left(0.066\right)\left(\pi {\left(1.25\times {10}^{-3}\right)}^2\right)\left(0.6707\right)}{2\left(1.68\times {10}^{-8}\right)}$

Q $=\text{5}\text{.23 C}$