Respuesta :
Answer:
(a)The average of magnetic flux trough each turns is [tex]3.63 \times 10^{-8}[/tex] wb.
(b) The magnitude of mutual inductance of the two solenoid is [tex]1.596 \times 10^{-5}[/tex] H.
(c)The magnitude of emf induced in the outer solenoid by changing current in the inner solenoid is 0.027132 v .
Explanation:
Given that,
The number of turns of the solenoid=N₁= 350
The diameter of the solenoid=D= 2.20 cm=0.022
The length of the solenoid is = [tex]l[/tex] = 22.0 cm=0.22 m
The second solenoid with [tex]N_2[/tex]= 21 turns which is wound around the solenoid at its center.
The current in the inner solenoid is [tex]I_1[/tex] = 0.150 A and is increasing at a rate of 1700 A/s i.e [tex]\frac{dI_1}{dt}= 1700[/tex] A/s
(a)
The formula of magnetic field of a solenoid is
[tex]B=\mu_0 nI=\frac{\mu_0 N_1I_1}{l}[/tex]
[tex]=\frac{(4\pi \times 10^{-7} T.m/A) (350)(0.15 \ A)}{0.220 \ m}[/tex]
[tex]\approx 3.0 \times 10^{-4}[/tex] T
So, the magnetic flux of each turns
[tex]\phi = BA=B \pi (\frac d2)^2[/tex]
[tex]=(3.0\times 10^{-4}\ T)\pi (\frac {0.022}{2} \ m)^2[/tex]
[tex]=1.14 \times 10^{-7}[/tex] wb
The average of magnetic flux trough each turns is [tex]3.63 \times 10^{-8}[/tex] wb.
(b)
Since the both coil wound tightly. So, the magnitude magnetic flux though each turns of both coils is same.
The mutual inductance of the two solenoid is
[tex]M=\frac{N_2\phi}{I_1}[/tex]
[tex]=\frac{(21)(1.14 \times 10^{-7}\ wb)}{0.15\ A}[/tex]
[tex]=1.596 \times 10^{-5}[/tex] H
(c)
The formula of emf is
[tex]\varepsilon =-M\frac{dI_1}{dt}[/tex]
[tex]=-(1.596\times 10^{-5}\ H) (1700 \ A/s)[/tex]
[tex]=27.132 \times 10^{-3}[/tex] v
=0.027132 v
The emf induced in the outer solenoid by changing current in the inner solenoid is 0.027132 v .
