A solenoidal coil with 21 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 22.0 cm long and has a diameter of 2.20 cm. At a certain time, the current in the inner solenoid is 0.150 A and is increasing at a rate of 1700 A/s. a. For this time, calculate the average magnetic flux through each turn of the inner solenoid.
b. For this time, calculate the mutual inductance of the two solenoids;
c. For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

(a)The average of magnetic flux trough each turns is wb.

(b) The magnitude of  mutual inductance of the two solenoid is 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 = = 22.0 cm=0.22 m

The second solenoid with = 21 turns which is wound around the solenoid at its center.

The current in the inner solenoid is = 0.150 A and is increasing at a rate of 1700 A/s i.e A/s

(a)

The formula of magnetic field of a solenoid is

                T

So, the magnetic flux of each turns

               wb

The average of magnetic flux trough each turns is 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

    H

(c)

The formula of emf is

  v        

 =0.027132 v

 The emf induced in the outer solenoid by changing  current in the inner solenoid is 0.027132 v .    

massxgravity x heght=weightxheight= pe

4900=490xh

h=10