Contents
Physics Topics cover a broad range of concepts that are essential to understanding the natural world.
What do you Mean by Non-Inductive Resistance Coil? What is a Choke?
When current flows through a coil, a magnetic field is generated around the coil [Fig.]. As a result, the coil becomes linked up with its own magnetic flux. This magnetic flux increases or
decreases with the increase or decrease of current. Hence, an electromagnetic induction takes place and an emf is induced in the coil. This induced emf opposes the change in electric current through the coil. This phenomenon is known as self-induction.
Definition: An electromagnetic induction in a coil due to change of current through itself is called self-induction.
Electric current generated due to self-induction flows in the direction opposite to the main current.
Self-inductance: If current flows through a coil, the magnetic flux linked with the coil becomes directly proportional to the current passing through it. So, if the magnetic flux is ϕ for current I, then
ϕ ∝ I or, ∝ = LI …. (1)
L is a constant which depends on its own construction. It is called the self-inductance or coefficient of self-induction of the coil. If I = 1, then L = ϕ
Definition: For unit current flowing through a coil, the mag-netic flux linked with it is called its self-inductance.
Again, from section 1.4.3, we know that the emf induced in the coil, e = \(-\frac{d \phi}{d t}\).
So, using equation (1) we can write,
e = -L\(\frac{d I}{d t}\) …. (2)
If \(\frac{d I}{d t}\) = 1, then e = L (considering the magnitude only). From this, the following alternative definition of self-inductance can be drawn:
If the rate of change of electric current in a coil with time be unity, the emf induced in the coil is called its self-inductance.
Unit of self-inductance:
In SI: In this system, the unit of self-inductance is henry (H). To define it, we can use either of the equations (1) or (2).
i) If I = 1 A and ϕ = 1 Wb , then L = 1 H.
ii) If \(\frac{d I}{d t}\) = 1 A ᐧ s-1 and e = 1 V, then L = 1 H .
So, self-inductance of a coil is 1 H if,
i) for a current of 1 A flowing through that coil, the magnetic flux linked with it is 1 Wb; or,
ii) for a change of 1A current in 1 second through that coil, the emf induced in it is 1 V.
Hence, from equation (1) we get,
Wb = H ᐧ A ….. (3)
Again from equation (2) we get,
V = H ᐧ \(\frac{\mathrm{A}}{\mathrm{s}}\) or, \(\frac{\mathrm{V}}{\mathrm{A}}\) ᐧ s = H
∴ H = Ω ᐧ s …….. (4)
The unit of resistance R is Ω and the unit of self-inductance L is H. Hence from equation (4) we infer that, the unit of \(\frac{L}{R}\) and the unit of time are the same.
In CGS System: Putting \(\frac{d I}{d t}\) = 1 emu ᐧ s-1 and e = 1 emu of emf, we get, L = 1 emu of self-inductance. So, if current be changed at the rate of 1 emu ᐧ s-1 and 1 emu electromotive force be induced in the coil, the self-inductance of that coil is 1 emu.
Relation between the CGS and SI units of self-inductance:
1 H = \(\frac{1 \mathrm{~V}}{1 \mathrm{~A} \cdot \mathrm{s}^{-1}}\) = \(\frac{10^8 \text { emu of potential difference }}{\frac{1}{10} \text { emu of current per second }}\)
Non-inductive coil: Self-induction creates disturbances in many electrical circuits. A coil of finite resistance but of zero self-inductance is often required. Such a coil is called a non-inductive coil.
A long insulated conducting wire is given a fold and then coiled up [Fig], The free ends of the coil are now on the same side. This kind of winding is called non-inductive winding. Hence antiparallel currents flow through adjacent wires of any part of the coil. So, the resultant magnetic flux becomes zero. That means, in this coil, no electromagnetic induction takes place.
Effect of self-induction in an electrical circuit: In the circuit shown in Fig., if the emf of the battery B be E, the effective emf of the circuit due to self-induction [according to equation (2)] = E-L\(\frac{d I}{d t}\).
If the current be zero, or a steady current flows through the circuit, \(\frac{d I}{d t}\) = 0; and in this case, self-induction has no effect on the circuit. On the other hand, self-induction affects a circuit when electric current changes with time in that circuit. Thus, self-induction plays an important role in an ac circuit. In case of a dc circuit, self-induction occurs only when the circuit is switched ‘on’ or ‘off!
Choke: A coil of high self-inductance and low resistance is commonly called a choke. If a choke is connected in series in an ac circuit, then due to its high self-inductance, the effective resistance of the circuit increases. On the other hand, due to low resistance of the coil, less heat is produced in the circuit. So, the function of a choke is to increase the effective resistance of the circuit lowering the dissipation of heat energy as much as possible.
Uses: In a tube-light arrangement, a choke is usually connected in series.
Energy stored in the magnetic field of an inductor: We have, induced emf in an inductor carrying a time varying current i, e = -L\(\frac{d i}{d t}\), where L = self-inductance of the inductive coil.
To increase the current in the inductor against this opposing emf, some energy has to be spent as work. This external work done will be stored in the magnetic field of the inductor as magnetic energy.
Power, i.e., rate of work done = ei = Li\(\frac{d i}{d t}\)
(the negative sign is not taken in calculation of work done)
So, work done in time dt = ei dt = Lidi
Total work done to increase the current through the inductor from 0 to I,
W = \(L \int_0^I i d i\) = \(L \cdot\left[\frac{i^2}{2}\right]_0^I\) = \(\frac{1}{2} L I^2\)
This work done is stored as energy (EL) in the magnetic field of the inductor, i.e.,
EL = \(\frac{1}{2} L I^2\) …. (5)
Mutual Induction
Let two coils, P and S, be kept very close to each other [Fig.]. Now if current is passed through one of them (say, P), a magnetic field is generated around it.
As a result, the second coil (S) gets linked with the magnetic flux generated by the first coil. If the current through P is increased or decreased with respect to time, the magnetic flux linked with an electromotive force is induced in the coil S will also increase or decrease accordingly. Hence, an electromotive force is induced in the coil S. This phenomenon is known as mutual induction. The first coil P is called the primary coil and the second coil S is called the secondary coil.
Definition: If current changes with respect to time in a coil, the electromagnetic induction that occurs in an adjacent coil is called mutual induction.
Mutual inductance: Mutual inductance is defined in a similar way as self-inductance. If the magnetic flux linked with the secondary coil be ϕs for current Ip in the primary coil then,
ϕS ∝ Ip or, ϕS = MIp ….. (1)
Here, M is a constant, which depends on the geometry of the coils and the distance between them and is called mutual inductance or coefficient of mutual induction.
Naturally for Ip = 1, M = ϕS
Definition: In case of two adjacent coils, if one carries a unit current, then the magnetic flux linked with the other is called the mutual inductance between the two coils.
Again, induced emf in the secondary coil,
e = \(-\frac{d \phi_S}{d t}\) = \(-M \frac{d I_P}{d t}\) …. (2)
If \(\frac{d I_P}{d t}\) = 1, then e = M [considering the magnitude only]
From this, the following alternative definition of mutual inductance is obtained.
In case of two adjacent coils, if the rate of change of current with respect to time in a coil be unity, the emf induced in the other coil is called the mutual inductance between the two coils.
Unit of mutual inductance:
In SI: In this system, the unit of mutual inductance is henry (H), exactly that of self inductance L.
‘Mutual inductance between a pair of coils is 1 H’ means that,
- if the current passing through one coil be 1 A, the magnetic flux linked with the other becomes 1 Wb; or,
- if the current changes through one coil at the rate of 1 A per second, the emf induced in the other becomes 1 V.
In CGS system: If \(\frac{d I}{d t}\) = 1 emu of current per second and e = 1 emu of potential, then M = 1 emu of mutual inductance. So, in case of two adjacent coils, if the current changes at the rate of 1 emu ᐧ s-1 in one coil and as a result if 1 emu of emf is induced in the other coil, the mutual inductance between the coils is said to be 1 emu.
Discussions:
Interchange of two coils: The second coil can be used as primary coil and the first as the secondary. In that case, due to change in current in the second coil, electromagnetic induction will take place in the first coil, and the value of mutual inductance (M) will remain unchanged.
Relation between self-inductance and mutual inductance: If the self-inductances of the two coils are L1 and L2, the mutual inductance becomes
M = \(k \sqrt{L_1 L_2}\) …….. (3)
Here, k is a constant whose value is 1 or less than 1. If the magnetic flux of one coil be linked completely with the other, k ≈ 1. Thus,
M ≈ \(\sqrt{L_1 L_2}\)
Inductance of a Solenoid
Self-inductance of a solenoid: Let the length of a solenoid = l, area of its circular cross section = A .
So, if number of turns be N, then the number of turns per unit length (n) = \(\frac{N}{l}\).
If current I flows through the solenoid, then the magnetic field thus generated inside it is
B = µ0nI, where µ0 = magnetic permeability of vacuum or air.
So, the magnetic flux linked with each turn of the solenoid,
BA = µ0nAI = \(\frac{\mu_0 N A}{l} \cdot I\)
Hence, the magnetic flux linked with N turns,
ϕ = NBA = \(\frac{\mu_0 N^2 A}{l} \cdot I\)
If the self-inductance of the solenoid be L then, ϕ = LI
∴ L = \(\frac{\phi}{I}\) = \(\frac{\mu_0 N^2 A}{l}\) …. (1)
For any other medium of magnetic permeability ji inside the solenoid, L = \(\frac{\mu N^2 A}{l}\).
Mutual inductance of two inseparable solenoids:
PP’ and SS’ are primary and secondary solenoids respectively [Fig.]. They are wound over each other on a length l.
Cross sectional area of each of them = A
Number of turns in length l in PP’ = N1, and that in SS’ = N2.
Therefore, number of turns per unit length of PP’,
n1 = \(\frac{N_1}{l}\)
If I1 current flows through the primary coil PP’, the magnetic field thus developed inside it,
B1 = µ0n1I1 = \(\frac{\mu_0 N_1}{l} I_1\)
[µ0 = magnetic permeability of vacuum or air]
As the solenoids are closely wound on each other, the magnetic flux linked in each turn of the secondary coil = B1A and hence the magnetic flux linked with N2 turns,
ϕ2 = N2B1A = \(\frac{\mu_0 N_1 N_2 A}{l} \cdot I_1\)
If the mutual inductance of the solenoid be M then,
ϕ2 = MI1
So, M = \(\frac{\mu_0 N_1 N_2 A}{l}\)
For any other medium used as core, M = \(\frac{\mu N_1 N_2 A}{l}\), where µ = magnetic permeability of that medium.
In the present context, self inductances of the coils are
Comparing this equation with equation (3) of section 1.7.2 we get, k = 1. This is the maximum value of k.
Energy density in magnetic field of a solenoid:
Length of a solenoid = A
cross sectional area = A
So, its volume = lA
If N be the total turns in the solenoid, the number of turns per unit length of it, n = \(\frac{N}{l}\).
If I current flows through the solenoid, the magnetic field thus developed along its axis,
B = µ0nI = \(\frac{\mu_0 N I}{l}\) or, I = \(\frac{B l}{\mu_0 N}\)
But, self-inductance of the solenoid,
L = \(\frac{\mu_0 N^2 A}{l}\)
So, the stored energy in its magnetic field,
U = \(\frac{1}{2} L I^2\) = \(\frac{1}{2} \frac{\mu_0 N^2 A}{l}\left(\frac{B l}{\mu_0 N}\right)^2\) = \(\frac{1}{2 \mu_0} B^2 l A\)
So, the energy stored per unit volume,
u = \(\frac{U}{l A}\)
i.e., u = \(\frac{1}{2 \mu_0} B^2\) … (3)
This u is the energy density in the magnetic field. For any other medium, instead of air or vacuum,
u = \(\frac{1}{2 \mu} B^2\)
Though equation (3) has been established for the magnetic field inside a solenoid, it is a general equation for any magnetic field because the relation is independent of the geometrical features of the inductor. To find the energy density at a point in any magnetic field the above equation may be applied.
Unit of energy density = J ᐧ m-3 and its dimension = \(\frac{M L^2 T^{-2}}{L^3}\) = ML-1T-2
Numerical Examples
Example 1.
The mutual inductance between two adjacent coils is 1.5 H. If the current in the primary coil changes from 0 to 20 A in 0.05 s, determine the average emf induced in the secondary coil. If the number of turns in the secondary coil is 800, what change in flux will be observed in it?
Solution:
The average value of induced emf,
e = \(-M \frac{\Delta I}{\Delta t}\) = -1.5 × \(\frac{(20-0)}{0.05}\) = \(-\frac{1.5 \times 20}{0.05}\) = -600 V
Hence, the magnitude of the induced emf = 600 V.
Again, from the relation ϕ = MI we get,
change in flux, Δϕ = M ᐧ ΔI = 1.5 × (20 – 0) = 30 Wb.
Example 2.
When current in a coil changes from + 2 A to -2 A in 0.05 s, an emf of 8 V is induced in the coil. Determine the self-inductance of the coil. [AIEEE 03]
Solution:
We know that, e = \(-L \frac{\Delta I}{\Delta t}\)
∴ Self-inductance of the coil, L = \(-\frac{e \Delta t}{\Delta I}\) = \(-\frac{8 \times 0.05}{-2-(+2)}\) = 0.1 H
Example 3.
Mutual inductance of two coils is 0.005H, ac in primary coil, I = I0sin ωt, where IQ = 10A and CO = 100π rad/s. What is the maximum emf in the secondary coil?
Solution:
ac in primary coil, I = I0sinωt
∴ \(\frac{d I}{d t}\) = ωI0 cosωt
S0, e2 = -M\(\frac{d I}{d t}\)[here M = mutual inductance]
= -MωI0 cosωt
So the maximum emf in the secondary coil
= MωI0 = 0.005 × 100 π × 10 = 15.7V
Example 4.
If a rate of change of current of 2 A ᐧ s-1 induces an emf of 10 mV in a solenoid, what is the self-inductance of the solenoid? [WBCHSE Sample Question]
Solution:
e = -L\(\frac{d I}{d t}\) or, L = \(\frac{|e|}{\frac{d I}{d t}}\) = \(\frac{10 \times 10^{-3} V}{2 \mathrm{~A} \cdot \mathrm{s}^{-1}}\) = 5 mH
Example 5.
Resistance of a coil is 10Ω and its self inductance is 5H. Find the energy stored when it is connected with a 100V battery.
Solution:
Current in circuit, I = \(\frac{V}{R}\) = \(\frac{100}{10}\) = 10A
Stored energy in the inductor = \(\frac{1}{2} L I^2\) = \(\frac{1}{2}\) × 5 × (10)2 = 250 J
Example 6.
Self-inductance of an air core solenoid increases from 0.01 mH to 10 mH when an iron core is introduced in it. What is the relative magnetic permeability of iron?
Solution:
Self-inductance is proportional to the permeability of the core. So, we get,
\(\frac{L}{L_0}\) = \(\frac{\mu}{\mu_0}\) = µr or µr = \(\frac{L}{L_0}\) = \(\frac{10}{0.01}\)
∴ µr = 1000
Hence relative magnetic permeability of iron is 1000 .
Example 7.
A small square loop of wire of side y is placed inside a large square loop of side x (x\(\gg\)y). The loops are coplanar and their centres coincide. Find the mutual inductance of the system.
Solution:
Let I be the current flowing through a square loop of side L.
Magnetic field at the centre of the loop, B = 4B1
[where B1 is magnetic field at the centre of the loop due to one of its sides]
Now, for the large square loop, L = x
Now, magnetic flux linked with the small square loop,
ϕ = B × area of the small square loop
= B × y2 = \(\frac{2 \sqrt{2} \mu_0 I y^2}{\pi x}\) …. (1)
If M be the mutual inductance between the two loops, then
ϕ = MI …. (2)
From equations (1) and (2) M = \(\frac{2 \sqrt{2} \mu_0 y^2}{\pi x}\)
Example 8.
Cross sectional area of a solenoid is 10 cm2. Half of its cross section is filled with Iron (µr = 450) and the remaining half with air (µr = 1). Calculate the self inductance of the solenoid if its length is 2 m and number of turns is 3000.
Solution:
If α1 part of cross section of the solenoid is filled with a substance of relative permeability \(\mu_{r_1}\) and the remaining part of cross section α2 with another substance of relative permeability \(\mu_{r_2}\) then self-inductance of the solenoid is,
L = \(\frac{\mu_0 n^2 A}{l}\left(\mu_{r_1} \alpha_1+\mu_{r_2} \alpha_2\right)\)
Here µ0 = 4π × 10-7 H/m; number of turns, n = 3000; length
of solenoid, l = 2 m; area of cross section, A = 10 cm2 = 0.001 m2; a1 = 0.5 and a2 = 0.5.
∴ Self-inductance of the solenoid,
L = \(\frac{4 \pi \times 10^{-7} \times(3000)^2 \times 0.001}{2}\)(1 × 0.5 + 450 × 0.5) H
= 2π × 9 × 10-4 × (0.5 + 225) H
= 1.27H (approx.)