Respuesta :
Answer:
0.0226 T, 7.096 V
Explanation:
N = 600
l = 10 cm = 0.1 m
i = 3 A
diameter = 2 cm
The magnetic field on the axis is given by
B = μo x n x i
where, n be the number of turns per unit length, i be the current and μo be the permeability of free space.
[tex]n =\frac{N}{l}= \frac{600}{0.1}=6000[/tex]
[tex]B=4\times 3.14 \times 10^{-7}\times 6000\times 3[/tex]
B = 0.0226 T
time taken to decrease the magnetic field to be zero is t = 0.001 second
Let e be the amount of voltage induced.
According to the Faraday's law of electromagnetic induction
e = rate of change of magnetic flux
[tex]e=\frac{d\phi }{dt}=A\frac{dB}{dt}[/tex]
where, A be the area of coil
[tex]A = 3.14 \times r^{2}= 3.14 \times 0.01\times 0.01=3.14\times 10^{-4}m^{2}[/tex]
So, induced voltage
[tex]e=\frac{3.14\times 10^{-4}\times 0.0226 }{0.001}=7.096\times 10^{-3}V\\[/tex]
e = 7.096 mV
Answer:
0.0226 T 4.25784 volt
Explanation:
Length of the solenoid = 10 cm =0.1 m
Diameter of the solenoid d=2 cm , so radius [tex]r=\frac{d}{2}=\frac{2}{1}=1cm=0.01m[/tex]
Number of turns N = 600, so number of turns per unit length [tex]n=\frac{N}{L}=\frac{600}{0.1}=6000[/tex]
Current = 3 A
Magnetic field due to solenoid [tex]B=\mu _0ni=4\pi \times 10^{-7}\times 6000\times 3=0.02261T[/tex]
Area [tex]A=\pi r^2=3.14\times 0.01^2=3.14\times 10^{-4}m^2[/tex]
Emf induced is given by [tex]e=-NA\frac{dB}{dt}=-600\times 3.14\times 10^{-4}\frac{(0-0226)}{0.001}=4.25784V[/tex]
