Physics II A solenoidal coil with 24 turns of wire is wound tightly around anoth
ID: 1511460 • Letter: P
Question
Physics II
A solenoidal coil with 24 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 23.0 cm long and has a diameter of 2.40 cm . At a certain time, the current in the inner solenoid is 0.110 A and is increasing at a rate of 1500 A/s .
----------------------------------------------------------------------------------------------------------------------------------------
Calculate the average magnetic flux through each turn of the inner solenoid. (Answer in Wb)
For this time, calculate the mutual inductance of the two solenoids; (Answer in H)
For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.(Answer in V)
Explanation / Answer
Number of turns in inner solenoid = N1 = 300
Number of turns in outer solenoid = N2 = 24
Length of inner solenoid = L=23.0 cm = 0.23 m
Radius of inner solenoid = r=diameter/2= 2.40/2=1.20 cm=0.012 m
The current in the inner solenoid = i = 0.110 A
Rate of increasing of current =di/dt = 1500 A/s.
The average magnetic flux through each turn of the inner solenoid. =B.A=(mu not)(N/L)i.A
The average magnetic flux through each turn of the inner solenoid. =[4(pi)*10^-7][300/0.23][0.11](pi)(0.012)...
A)The average magnetic flux through each turn of the inner solenoid. =6.79*10^-6 weber
_____________________________
The mutual inductance of the two solenoids=N2* 6.79*10^-6
(B) The mutual inductance of the two solenoids =1.62*10^-4 weber/A