Question

A solenoid of length 20 cm, area of cross-section 4.0 cm2 and having 4000 turns is placed inside another solenoid of 2000 turns having a cross-sectional area 8.0 cm2 and length 10 cm. Find the mutual inductance between the solenoids.

Answer 1

Answer:

cross section of open centimetre but it can be because a solenoid is of only 2000 turns not 2012 if it is placed in it is a special introduction

Answer 2

The mutual inductance between the solenoids is  $2 \times 10^{-2} H$

Explanation:

Step 1:

Mutual inductance

$M=\mu_{0} N_{1} N_{2} \pi r_{1}^{2} l$

$\begin{array}{l}{=4 \pi \times 10^{-7} \times 4 \times 10^{3} \times 2 \times 10^{3} \times 4 \times 10^{-4} \times 10 \times 10^{-2}} \\{=0.04 \times 10^{-2} H}\end{array}$

Step 2:

In case of  solenoid-1

Cross section area, a1 = 4 cm2 = $4 \times 10^{-4} \mathrm{m}^{2}$

Solenoid length, l1 = 20 cm = 0.20 m

No of turns or moves per unit length ,$n_1$ =4000/0.2  m = 20000 turns/m

In case of  solenoid-2

Cross section area, a2 = 8 cm2 = 4 × 10−4 m2

Solenoid length, $l_2$ = 10 cm = 0.1 m

no. of moves or turns per unit length, $n_2$ = 2000/0.1 m = 20000 turns/m

The solenoid-1 is given to be inside the solenoid-2

Let the present be I by the solenoid-2

In solenoid-2 the magnetic field due to current

$B=\mu_{0} n_{2} i=4 \pi \times 10^{-7} \times 20000 \times i$

Step 3:

Flux is given via coil-1

$\emptyset=n_{1} l_{1}, B . a_{1}=n_{1} l_{1}, \mu_{0} n_{2} i \times a 1$

$\phi=2000 \times 20000 \times 4 \pi \times 10^{-7} \times i \times 4 \times 10^{-4}$

If the current flow in the coil-2 changes, then emf in the coil-1 is induced

Thus, emf induced in the coil-1 is given because of the current change in the coil-2

$E=\frac{d \emptyset}{d t}=64 \pi \times 10^{-4} \times \frac{d i}{d t}$

$M=\frac{E}{\frac{d i}{d t}}=64 \pi \times 10^{-4} H=2 \times 10^{-2} H \quad\left(E=M \frac{d i}{d t}\right)$