Answer:
a) 1.34*10^-8 W
b) 1.18*10^-5 H
c) 20mV
Explanation:
a) To find the average magnetic flux trough the inner solenoid you the following formula:
[tex]\Phi_B=BA=\mu_oNIA[/tex]
mu_o: magnetic permeability of vacuum = 4pi*10^-7 T/A
N: turns of the solenoid = 340
I: current of the inner solenoid = 0.100A
A: area of the inner solenoid = pi*r^2
r: radius of the inner solenoid = 2.00cm/2=1.00cm=10^-2m
You calculate the area and then replace the values of N, I, mu_o and A to find the magnetic flux:
[tex]A=\pi(10^{-2}m)^2=3.141510^{-4}m^2\\\Phi_B=(4\pi*10^{-7}T/A)(340)(0.100A)(3.1415*10^{-4}m^2)=1.34*10^{-8}W\\[/tex]
the magnetic flux is 1.34*10^{-8}W
b) the mutual inductance is given by:
[tex]M=\mu_o N_1 N_2 \frac{A_2}{l}[/tex]
N1: turns of the outer solenoid = 22
N2: turns of the inner solenoid
A_2: area of the inner solenoid
l: length of the solenoids = 25.0cm=0.25m
by replacing all these values you obtain:
[tex]M=(4\pi*10^{-7}T/A)(340)(22)\frac{3.14*10^{-4}m^2}{0.25m}=1.18*10^{-5}H[/tex]
the mutual inductance is 1.18*10^{-5}H
c) the emf induced can be computed by using the mutual inductance and the change in the current of the inner solenoid:
[tex]\epsilon_1=M\frac{dI_2}{dt}[/tex]
by replacing you obtain:
[tex]\epsilon_1=(1.18*10^{-5}H)(1700A/s)=0.02V=20mV[/tex]
the emf is 20mV
A) The average magnetic flux through each turn of the inner solenoid is ; 1.34 * 10⁻⁸ W
B) The mutual inductance of the two solenoids = 1.18 * 10⁻⁵ H
C) Induced emf in the outer solenoid = 20 mV
Given data :
Number of inner solenoid turns ( N₂ ) = 340 turns
( N₁ ) outer solenoid turns = 22 turns
Length of inner solenoid = 25.0 cm = 0.25 m
Diameter of Inner solenoid = 2.00 cm ∴ radius = 10⁻²m
Current in solenoid ( I ) = 0.10 A
Rate of change of current in Inner solenoid ( r ) = 1700 A/s
Area = π*r² = π * ( 10⁻² )² = 3.14151 * 10⁻⁴
A) Determine the average magnetic flux through each turn
Average magnetic flux ( Ф B ) = μo*N*I*A ----- ( 1 )
where ; μo = 4π* 10⁻⁷ T/A , N = 340, I = 0.10, A = 3.141510 * 10⁻⁴
Insert values into equation ( 1 )
Average magnetic flux ( ФB ) = 1.34 * 10⁻⁸ W
B) Calculate the Mutual inductance
M = μo N₁N₂[tex]\frac{A_{2} }{l}[/tex] ------ ( 2 )
where ; μo = 4π* 10⁻⁷, N₁ = 22, N₂ = 340, A₂ = 3.14151 * 10⁻⁴, l = 0.25
Insert values into equation ( 2 )
Mutual inductance ( M ) = 1.18 * 10⁻⁵ H
C ) Calculate the Induced emf in the outer solenoid
Induced emf = M * r
= 1.18 * 10⁻⁵ * ( 1700 ) = 0.02V ≈ 20 mV
Hence we can conclude that The average magnetic flux through each turn of the inner solenoid is ; 1.34 * 10⁻⁸ W, The mutual inductance of the two solenoids = 1.18 * 10⁻⁵ H and Induced emf in the outer solenoid = 20 mV.
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