Chapter 20, Problem 20QAP

To determine

(a)

Induced emf at t < 0 s

Answer to Problem 20QAP

Induced emf at t < 0 s= $0\text{ T}$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Number of turns =30}\\ \text{diameter of the coil =0}\text{.6 m}\\ \text{magnetic field = }1.0\text{ T}\end{array}$

Formula used:

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

Calculation:

  $radius=\frac{0.6\text{ m}}{2}=0.3\text{ m}$

  $\begin{array}{l}\text{By substituting,}\\ \epsilon \text{=}\frac{-d\Phi }{dt}\\ \epsilon \text{=}\frac{-d}{dt}(NB\pi r^2)\\ \epsilon \text{=N}\pi r^2\frac{-dB}{dt}\\ \epsilon \text{=-N}\pi r^2\frac{(B_2-B_1)}{dt}\\ \text{Magnetic field is constant when t < 0 s}\\ dB\text{ = 0,}\\ \epsilon =0\end{array}$

Conclusion:

Induced emf at t < 0 s= $0\text{ T}$

To determine

(b)

Induced emf at t =5.0 s

Answer to Problem 20QAP

Induced emf at t =5.0 s= $2.54\text{ mV}$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Number of turns =30}\\ \text{diameter of the coil =0}\text{.06 m}\\ \text{magnetic field = }1.0\text{ T}\\ \text{magnetic field is increased to 1}\text{.3 T in 10 s}\end{array}$

Formula used:

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

Calculation:

  $\begin{array}{l}radius=\frac{0.06\text{ m}}{2}=0.03\text{ m}\\ \text{Magnetic field at 5 s = }\frac{\text{1}\text{.3 T - 1}\text{.0 T}}{10\text{ s}}*5\text{ s+1}\text{.0 T=1}\text{.15 T}\end{array}$

  $\begin{array}{l}\text{By substituting,}\\ \epsilon \text{=}\frac{-d\Phi }{dt}\\ \epsilon \text{=}\frac{-d}{dt}(NB\pi r^2)\\ \epsilon \text{=N}\pi r^2\frac{-dB}{dt}\\ \epsilon \text{=-N}\pi r^2\frac{(1.15\text{ T}-1.00\text{ T})}{dt}\\ \epsilon \text{=-30*}3.14*{(}^{0.03}\frac{(1.15\text{ T}-1.00\text{ T})}{1\text{ s}}\\ \epsilon =-2.54\text{ mV}\end{array}$

Conclusion:

Induced emf at t =5.0 s= $2.54\text{ mV}$

To determine

(c)

Induced emf at t < 10 s

Answer to Problem 20QAP

Induced emf at t < 10 s= $0\text{ T}$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Number of turns =30}\\ \text{diameter of the coil =0}\text{.6 m}\\ \text{magnetic field = }1.0\text{ T}\end{array}$

Formula used:

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

Calculation:

  $radius=\frac{0.6\text{ m}}{2}=0.3\text{ m}$

  $\begin{array}{l}\text{By substituting,}\\ \epsilon \text{=}\frac{-d\Phi }{dt}\\ \epsilon \text{=}\frac{-d}{dt}(NB\pi r^2)\\ \epsilon \text{=N}\pi r^2\frac{-dB}{dt}\\ \epsilon \text{=-N}\pi r^2\frac{(B_2-B_1)}{dt}\\ \text{Magnetic field is constant after 10 s}\\ dB\text{ = 0,}\\ \epsilon =0\end{array}$

Conclusion:

Induced emf at t < 10 s= $0\text{ T}$

To determine

(d)

Plot the magnetic field and induced emf as functions of time

Answer to Problem 20QAP

  

  

Explanation of Solution

Given info:

Time (s)	Magnetic field (T)	induced emf (V)
-10	1	0
-5	1	0
0	1	0
1	1.03	0.0025434
2	1.06	0.0025434
3	1.09	0.0025434
4	1.12	0.0025434
5	1.15	0.0025434
6	1.18	0.0025434
7	1.21	0.0025434
8	1.24	0.0025434
9	1.27	0.0025434
10	1.3	0.0025434
15	1.3	0
20	1.3	0

Formula used:

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

Calculation:

emf at 1 s as shown below, all the emf values were calculated like that by substituting the magnetic field at particular time.

  $\begin{array}{l}B\text{ at 1 s =1}\text{.03 T}\\ \text{By substituting,}\\ \epsilon \text{=}\frac{-d\Phi }{dt}\\ \epsilon \text{=}\frac{-d}{dt}(NB\pi r^2)\\ \epsilon \text{=N}\pi r^2\frac{-dB}{dt}\\ \epsilon \text{=-30*}3.14*{(}^{0.03}\frac{(1.03\text{ T}-1.00\text{ T})}{1\text{ s}}\\ \epsilon =2.54\text{ mV}\end{array}$

Conclusion:

Graphs were drawn in the answer section.