Question

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.

Answer 1

Answer:

(a)The average of magnetic flux trough each turns is $3.63 \times 10^{-8}$ wb.

(b) The magnitude of  mutual inductance of the two solenoid is $1.596 \times 10^{-5}$ 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 = $l$ = 22.0 cm=0.22 m

The second solenoid with $N_2$= 21 turns which is wound around the solenoid at its center.

The current in the inner solenoid is $I_1$ = 0.150 A and is increasing at a rate of 1700 A/s i.e $\frac{dI_1}{dt}= 1700$ A/s

(a)

The formula of magnetic field of a solenoid is

$B=\mu_0 nI=\frac{\mu_0 N_1I_1}{l}$

                 $=\frac{(4\pi \times 10^{-7} T.m/A) (350)(0.15 \ A)}{0.220 \ m}$

                $\approx 3.0 \times 10^{-4}$ T

So, the magnetic flux of each turns

$\phi = BA=B \pi (\frac d2)^2$

            $=(3.0\times 10^{-4}\ T)\pi (\frac {0.022}{2} \ m)^2$

             $=1.14 \times 10^{-7}$  wb

The average of magnetic flux trough each turns is $3.63 \times 10^{-8}$ 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

$M=\frac{N_2\phi}{I_1}$

     $=\frac{(21)(1.14 \times 10^{-7}\ wb)}{0.15\ A}$

    $=1.596 \times 10^{-5}$ H

(c)

The formula of emf is

$\varepsilon =-M\frac{dI_1}{dt}$

  $=-(1.596\times 10^{-5}\ H) (1700 \ A/s)$

  $=27.132 \times 10^{-3}$ v        

 =0.027132 v

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