Chapter 20, Problem 24QAP

To determine

(a)

Current through the wire coil before the coil reaches the edge of the field.

Answer to Problem 24QAP

Current through the wire coil before the coil reaches the edge of the field = $0$ A

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Number of turns =30}\\ \text{side length coil =0}\text{.1 m}\\ \text{side length coil =0}\text{.820 }\Omega \\ \text{magnetic field = 0}\text{.600 T}\\ \text{Field drops to 0 at the edges}\\ \text{Velocity of the coil=0}{\text{.02 ms}}^{\text{-1}}\text{ to right}\end{array}$

Formula used:

  $\begin{array}{l}A=d^2\\ A=\text{area}\\ d=\text{length of a side}\\ \Phi =NBA\\ N=\text{number of turns}\\ \epsilon \text{=}\frac{-d\Phi }{dt}\\ \epsilon =\text{ emf}\\ t=\text{ time}\end{array}$

Calculation:

  $Area={(0.1\text{ m)}}^{\text{2}}=0.01{\text{ m}}^2$

  $\begin{array}{l}\epsilon \text{=}\frac{-d\Phi }{dt}\\ \epsilon \text{=}\frac{-d}{dt}(NB{l}^2)\\ \epsilon \text{=N}{l}^2\frac{-dB}{dt}\\ \epsilon \text{=-N}{l}^2\frac{(B_2-B_1)}{dt}\end{array}$

  $\begin{array}{l}\text{There is no magnetic field difference, coil }\exp \text{erience the same }B\text{ when moving}\\ dB\text{ = 0,}\\ \epsilon =0\\ \text{there is no induced current}\end{array}$

Conclusion:

Current through the wire coil before the coil reaches the edge of the field = $0$ A

To determine

(b)

Current through the wire coil while leaving the magnetic field.

Answer to Problem 24QAP

Current through the wire coil while leaving the magnetic field =

  $0.044\text{ A}$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Number of turns =30}\\ \text{side length coil =0}\text{.1 m}\\ \text{side length coil =0}\text{.820 }\Omega \\ \text{magnetic field = 0}\text{.600 T}\\ \text{Field drops to 0 at the edges}\\ \text{Velocity of the coil=0}{\text{.02 ms}}^{\text{-1}}\text{ to right}\end{array}$

Formula used:

  $\begin{array}{l}A=l^2\\ A=\text{area}\\ l=\text{length of a side}\\ \Phi =NBA\\ \Phi =\text{magnetic flux}\\ N=\text{number of turns}\\ Q\text{=}\frac{Nd\Phi }{R}\\ Q=\text{ emf}\\ t=\text{ time}\\ V=IR\\ V=\text{voltage}\\ I=\text{current}\\ R=\text{resistance}\\ \end{array}$

Calculation:

  $Area={(0.1\text{ m)}}^{\text{2}}=0.01{\text{ m}}^2$

  $\begin{array}{l}\epsilon \text{=}\frac{-d\Phi }{dt}\\ \epsilon \text{=}\frac{-d}{dt}(NBA)\\ \epsilon \text{=N}A\frac{-dB}{dt}\\ A=l*l\\ \epsilon \text{=-N}*l^2\frac{(B_2-B_1)}{dt}\\ velocity=v=\frac{l}{dt}\\ \epsilon \text{=N}vl*0.6T\\ \epsilon =30*0.1\text{ m*}0.02{\text{ ms}}^{\text{-1}}\text{*0}\text{.6 T}\\ \epsilon =0.036\text{ V}\\ V=IR\\ I=\frac{V}{R}=\frac{\epsilon }{R}=\frac{0.036\text{ V}}{\text{0}\text{.820 }\Omega }=0.044\text{ A}\end{array}$

Conclusion:

Current through the wire coil while leaving the magnetic field = $0.044\text{ A}$

To determine

(c)

Current through the wire coil after leaving the magnetic field.

Answer to Problem 24QAP

Current through the wire coil after leaving the magnetic field= $0$ A

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Number of turns =30}\\ \text{side length coil =0}\text{.1 m}\\ \text{side length coil =0}\text{.820 }\Omega \\ \text{magnetic field = 0}\text{.600 T}\\ \text{Field drops to 0 at the edges}\\ \text{Velocity of the coil=0}{\text{.02 ms}}^{\text{-1}}\text{ to right}\end{array}$

Formula used:

  $\begin{array}{l}A=d^2\\ A=\text{area}\\ d=\text{length of a side}\\ \Phi =NBA\\ \Phi =\text{magnetic flux}\\ N=\text{number of turns}\\ \epsilon \text{=}\frac{-d\Phi }{dt}\\ \epsilon =\text{ emf}\\ t=\text{ time}\end{array}$

Calculation:

  $Area={(0.1\text{ m)}}^{\text{2}}=0.01{\text{ m}}^2$

  $\begin{array}{l}\epsilon \text{=}\frac{-d\Phi }{dt}\\ \epsilon \text{=}\frac{-d}{dt}(NB{l}^2)\\ \epsilon \text{=N}{l}^2\frac{-dB}{dt}\\ \epsilon \text{=-N}{l}^2\frac{(B_2-B_1)}{dt}\end{array}$

  $\begin{array}{l}\text{There is no magnetic field difference once it is left the magnetic field}\\ dB\text{ = 0,}\\ \epsilon =0\\ \text{there is no induced current}\end{array}$

Conclusion:

Current through the wire coil after leaving the magnetic field= $0$ A

To determine

(d)

Total charge that flows past a given point in the coil as it leaves the field

Answer to Problem 24QAP

Total charge that flows past a given point in the coil as it leaves the field= $0.22\text{ C}$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Number of turns =30}\\ \text{side length coil =0}\text{.1 m}\\ \text{side length coil =0}\text{.820 }\Omega \\ \text{magnetic field = 0}\text{.600 T}\\ \text{Field drops to 0 at the edges}\\ \text{Velocity of the coil=0}{\text{.02 ms}}^{\text{-1}}\text{ to right}\end{array}$

Formula used:

  $\begin{array}{l}A=d^2\\ A=\text{area}\\ d=\text{length of a side}\\ \Phi =NBA\\ \Phi =\text{magnetic flux}\\ N=\text{number of turns}\\ \epsilon \text{=}\frac{-d\Phi }{dt}\\ \epsilon =\text{ emf}\\ t=\text{ time}\\ \text{Q=}\frac{Nd\Phi }{R}\\ R=\text{resistance}\\ \Phi =BA\\ \text{Q=}\frac{30*(BA-0)}{R}\end{array}$

Calculation:

  $\begin{array}{l}\text{Q=}\frac{Nd\Phi }{R}\\ R=\text{resistance}\\ \Phi =BA\\ \text{Q=}\frac{30*(BA-0)}{R}\end{array}$

  $\begin{array}{l}\text{Q=}\frac{30*(0.6\text{ T*(0}{\text{.1 m)}}^{\text{2}}-0)}{0.82\Omega }\\ \text{Q}=0.22\text{ C}\end{array}$

Conclusion:

Total charge that flows past a given point in the coil as it leaves the field= $0.22\text{ C}$

To determine

(e)

Plot the induced current as a function of horizontal position

Answer to Problem 24QAP

  

Explanation of Solution

Conclusion:

Graph is plotted, induced current vs the horizontal position.