Consider two long co-axial solenoids of same length l.

**Mutual inductance**** ****between two long
co-axial solenoids**

Consider two long
co-axial solenoids of same length *l*. The length of these solenoids is
large when compared to their radii so that the magnetic field produced inside
the solenoids is uniform and the fringing effect at the ends may be ignored.
Let *A*_{1} and *A*_{2} be the area of cross section
of the solenoids with *A*_{1}* *being greater than* A*_{2}*
*as shown in Figure* *4.23. The turn density of these solenoids are *n*_{1}*
*and* n*_{2}* *respectively.

Let *i*_{1}
be the current flowing through solenoid 1, then the magnetic field produced
inside it is

As the field lines of _{1}
are passing through the area bounded by solenoid 2, the magnetic flux is linked
with each turn of solenoid 2 due to solenoid 1 and is given by

The flux linkage of
solenoid 2 with total turns *N*_{2} is

Comparing the equations
(4.20) and (4.21),

This gives the
expression for mutual inductance *M*_{21} of the solenoid 2 with
respect to solenoid 1. Similarly, we can find mutual inductance *M*_{12}
of solenoid 1 with respect to solenoid 2 as given below.

The magnetic field
produced by the solenoid 2 when carrying a current *i*_{2} is

*B*_{2}* *=* *µ* n*_{2}*i*_{2}

This magnetic field *B*_{2}
is uniform inside the solenoid 2 but outside the solenoid 2, it is almost zero.
Therefore for solenoid 1, the area *A*_{2} is the effective area
over which the magnetic field *B*_{2} is present; not area *A*_{1}.
Then the magnetic flux* Φ*_{12}* *linked with* *each
turn of solenoid 1 due to solenoid 2 is

The flux linkage of
solenoid 1 with total turns *N*_{1} is

From equation (4.22) and
(4.23), we can write

In general, the mutual
inductance between two long co-axial solenoids is given by

If a dielectric medium
of relative permeability µ* _{r}* is present inside the solenoids, then

M = µ n_{1} n_{2} A_{2}
l

(or) M = µ µ_{r}
n_{1} n_{2} A_{2} l

**EXAMPLE 4.12**

The current flowing in
the first coil changes from 2 A to 10 A in 0.4 sec. Find the mutual inductance
between two coils if an emf of 60 mV is induced in the second coil. Also
determine the induced emf in the second coil if the current in the first coil
is changed from 4 A to 16 A in 0.03 sec. Consider only the magnitude of induced
emf.

**Solution**

Case (i):

*di*_{1} = 10 – 2 = 8 A; d*t* =
0.4 s;

ε_{2} = 60 ×10^{−3}V

Case (ii):

di_{1} = 16 – 4
= 12 A;

dt = 0.03 s

(i) Mutual inductance of
the second coil with respect to the first coil

(ii) Induced emf in the
second coil due to the rate of change of current in the first coil is

**EXAMPLE 4.13**

Consider two coplanar,
co-axial circular coils A and B as shown in figure. The radius of coil A is 20
cm while that of coil B is 2 cm. The number of turns is 200 and 1000 for coils
A and B respectively. Calculate the mutual inductance of coil B with respect to
coil A. If the current in coil A changes from 2 A to 6 A in 0.04 sec, determine
the induced emf in coil B and the rate of change of flux through the coil B at
that instant.

**Solution**

N_{A} = 200
turns; N_{B} = 1000 turns;

r_{A} = 20 × 10^{-2}
m; r_{B} = 2 × 10^{-2} m;

dt = 0.04 s; di_{A}
= 6−2 = 4A

Let i_{A} be the
current flowing in coil A, then the magnetic field B_{A} at the centre
of the circular coil A is

The rate of change of magnetic flux of coil is

Tags : Definition, Explanation, Formulas, Solved Example Problems | Electromagnetic Induction , 12th Physics : Electromagnetic Induction and Alternating Current

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

12th Physics : Electromagnetic Induction and Alternating Current : Mutual inductance between two long co-axial solenoids | Definition, Explanation, Formulas, Solved Example Problems | Electromagnetic Induction

**Related Topics **

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.