Respuesta :
Answer:
a) B = 0.015 T
b) ФB = 4.71*10⁻6 W
c) emf = 7.89*10^-4 V
d) B = 60.31 T
Explanation:
(a) To find the magnitude of the magnetic field inside the solenoid you use the following formula:
[tex]B=\frac{\mu_oNI}{L}[/tex] (1)
B: magnitude of the magnetic field
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
N: turns of the solenoid = 300
L: length = 2.50cm = 0.025 m
I: current = 12 A
You replace the values of the variables in the equation (1):
[tex]B=\frac{(4\pi *10^{-7}T/A)(300)(12A)}{0.3m}=0.015T[/tex]
(b) The magnetic flux is given by:
[tex]\Phi_B=BA[/tex]
A: area = π(0.01m)^2 = 3.1415*10^-4 m^2
[tex]\Phi_B=(0.015T)(3.1415*10^{-4}m^2)=4.71*10^{-6}W[/tex]
(c) The induced emf is given by the following formula:
[tex]emf=-\frac{\Delta \Phi_B}{\Delta t}=\frac{A(B_f-B_i)}{\Delta t}[/tex] (2)
Bf and Bi are the final and initial magnetic field. You use the equation (1) into the equation (2):
[tex]emf=-A\mu_o\frac{N}{L}\frac{(I_2-I_1)}{\Delta t}\\\\emf=-\pi(0.01m)^2(4\pi*10^{-7}T/A)\frac{300}{0.3m}\frac{(10A-12A)}{0.001s}\\\\emf=7.89*10^{-4}V[/tex]
(d) With the ferromagnetic material inside the solenoid the magnetic field inside is modified as following:
[tex]B=\frac{\mu NI}{L}\\\\\mu=4000\mu_o=4000(4\pi*10^{-7}T/A)=5.026*10^{-3}T/A\\\\B=\frac{(5.026*10^{-3}T/A)(300)(12A)}{0.3m}=60.31T[/tex]
Following are the calculation to the given points:
Given:
[tex]r = 1.25\ cm \\\\l=30\ cm\\\\ N= 300 \\\\I=12\ A\\\\[/tex]
To find:
[tex]B=?\\\\\Phi_B=?\\\\[/tex]
Solution:
For point a:
Using formula:
[tex]\to \mu_o= 4 \pi 10^{-7} \ \frac{T}{A}[/tex]
[tex]\to N= 300\\\\ \to L= 2.50\ cm = 0.025\ m\\\\ \to I= 12 \ A\\\\[/tex]
[tex]\to B = \frac{\mu_o \ N\ I}{L}\\[/tex]
[tex]=\frac{4 \pi \times 10^{-7}\ \frac{T}{A} \times 300 \times 12\ A}{0.3\ m}\\\\=\frac{4 \times 3.14 \times 10^{-7}\ \frac{T}{A} \times 300 \times 12\ A}{0.3\ m}\\\\=\frac{48 \times 3.14\ T \times 300 }{0.3\ m \times 10^{7}}\\\\=\frac{ 150.72 \ T \times 3000 }{3\ m \times 10^{7}}\\\\=\frac{ 150.72 \ T \times 1000 }{10^{7}}\\\\= 0.015\ T\\\\[/tex]
For point b:
[tex]\to A : area = \pi (0.01m)^2 = 3.14 \times 10^{-4}\ m^2\\\\[/tex]
Using formula:
[tex]\to \Phi_B = BA\\\\[/tex]
[tex]= (0.015\ T)(3.14 \times 10^{-4} \ m^2)\\\\ = 4.71 \times 10^{-6}\ W[/tex]
For point c:
Using formula:
[tex]\to emf = -A \mu_o \frac{N}{L} \frac{(I_2-I_1)}{\Delta t}[/tex]
[tex]= -\pi(0.01 \ m)^2(4\pi \times 10^{-7}\ \frac{T}{A})\frac{300}{0.3\ m} \frac{(10A-12A)}{0.001\ s} \\\\= -3.14 (0.01 \times 0.01 \ m)(4\times 3.14 \times 10^{-7}\ \frac{T}{A})\frac{300}{0.3} \frac{(-2A)}{0.001\ s} \\\\= -3.14 (1 \ m)(12.56\times 10^{-7}\ \frac{T}{A}) 10^6 \frac{(-2A)}{ 10^3\ s} \\\\= -3.14 (1 \ m)(12.56 \frac{T}{A}) \frac{(-2A)}{ 10^4\ s} \\\\= 7.89 \times 10^{-4}\ V\\\\[/tex]
For point d:
[tex]\to B = \frac{\mu NI}{L}\\\\\to \mu = 4000 \mu_o = 4000(4\pi \times 10^{-7}\ \frac{T}{A}) = 5.026 \times 10^{-3} \frac{T}{A}\\\\ \to B = \frac{(5.026\times 10^{-3} \ \frac{T}{A})(300)(12\ A)}{0.3\ m} = 60.31\ T\\\\[/tex]
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