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Unit5(a): Magnetic forces, Materials and Inductance
Introduction: The main difference between static electric filed and static magnetic field is that electric field can exert a force on either stationary charge or moving charge, where as the steady magnetic field can exert force on moving charge. The study of these forces are important in understanding devices like ammeters, voltmeters, MHD generators, Plasmas, Galvanometers, relays, motors etc.
Force on a Moving Point Charge (Lorentz force equation)
The electrostatic force exerted on a point charge placed in an electric field can be given by coulomb’s law as 𝑭 e = Q 𝑬 (N)
If this charge moving with a constant velocity v placed in a steady magnetic field, it experiences a magnetic force given by 𝑭 𝒎 = 𝑄 𝒗 × 𝑩 (N)
The direction of 𝑭 𝒎 is perpendicular to the plane containing𝒗 𝑎𝑛𝑑 𝑩 . The electric force is independent of velocity where as the magnetic force is dependent on the velocity.
The total force exerted on a moving point charge in presence of both electric and magnetic fields can be given by 𝑭 = 𝑭 𝑻 𝒆+ 𝑭 𝒎
𝑭 = 𝑸[𝑬 + 𝑽 × 𝑩 ] --- (A)
Equation (A) is known as Lorentz-Force equation which relates the mechanical force to electrical force. If m is mass of the charge, then 𝑭 = m 𝒂 = 𝒎 𝒅𝒗𝒅𝒕 = 𝑄(𝑬 + 𝒗 × 𝑩 )
The solution to Lorentz –force equation is used in (i) determining the electron orbits in magnetron (ii) Proton paths in cyclotron (iii) Plasma characteristics in Magneto Hydro Dynamic Generators (MHDG) etc.
Note: Electric field and electric force acts along the same direction, where as the magnetic force acts perpendicular to both V and B vectors. Also electric force is exerted whether the charge is moving or stationary, where as the magnetic force will be exerted only when the charge is moving.
The magnitude of the static electric force is very small compared to magnetic force in a small region. Hall Effect:-
A differential element of charge is made up of very large number of small and discrete charges, occupies a volume which is larger than the average separation between the charges.
Consider conductor in which the electrons are in motion in the region of immobile +ve ions which form a crystalline array giving the conductor its solid properties.
A magnetic field which exerts forces on electrons tend to cause them to shift their positions and produces a small displacement between the centres of gravity of +ve and –ve charges. But this displacement is opposed by the columbic forces between the nucleus (+ve ions) and electrons. Any attempt to move the electrons therefore results in an attractive force between the +ve ions of the crystalline lattice and electrons. Hence the force is transferred to the crystalline lattice or to the conductor itself.
The columbic forces are very much greater than the magnetic forces in a good conductors and the displacement of electrons is almost immeasurable.
Other than perfect conductors, a separation of charges is observed which results in a small potential difference across the conductor, perpendicular to the magnetic field and to the velocity 𝒗 with which the charges are moving. This small voltage across the conductor is called Hall voltage and this effect is called Hall Effect.
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Hall effect is used to measure the Magnetic flux density.
Hall effect can be adapted in devices like electronic watt-meters, squaring elements where in current through the device is proportional to the magnetic field.
Hall effect is used to determine the given semiconductor material is p-type or n-type as the hall voltages for both the materials are different.
Force on a differential current Element
The differential force dF = dQ 𝐯 × 𝐁 (N) ---(1) (since Fe<<Fm) The current in a conductor can be expressed as I = Q / t
If the charge moves through a distance of l mts inside the conductor in t sec, then 𝒗 = 𝒍 𝒕 I l = Q( l / t ) = Q 𝒗
In the differential form, I.dl = dQ. 𝒗 ---(2) Combining equations (1) and (2)
(N) ---Differential form
The total force can be given by 𝐅 = (𝐈𝐝𝐥𝐥 𝐁 ) --- Integral form
For a straight long conductor of length l mts, carrying current I amps, the force exerted is given by the above equation as
𝑭 = 𝑩 Il sin = IL 𝑩 (N)
Where is the angle between the vectors representing the current and direction of the magnetic field.(Fleming’s left hand rule)
Note: By using the concept of Biot-Savart’s law applied to differential surfaces, we can write d𝐅 = I𝐝𝐥 × 𝐁 = 𝐉 𝐝𝐯 × 𝐁 = 𝐊 𝐝𝐬 × 𝐁 --- differential form
𝐅 = 𝐈 𝐝𝐥𝐥 × 𝐁 = 𝐊 × 𝐁 𝐝𝐬 = 𝐉 × 𝐁 𝐝𝐯 --- Integral form Force between two current elements
Consider two current carrying elements I1𝐝𝐥 1 and I2𝐝𝐥 2 as shown in the fig. Both currents produce their own
magnetic fields. The differential differential force exerted on current element I1𝐝𝐥 1 due to magnetic field 𝐝𝐁 2 produced by the current in the other element I2𝐝𝐥 2 is given by
d(d𝑭 1) = I1𝐝𝐥 1 𝐝𝐁 2 ---(1)
from BSL 𝐝𝐁 = 𝜇𝑑𝑯 2 =𝜇 𝐼4𝜋 𝑹2𝒅𝒍 ×𝒂 2 𝑹𝟐𝟏
𝟐𝟏2
using this value we get
𝑑 𝑑𝑭 = 𝜇𝐼𝟏 1𝐝𝐥 1×𝜇 𝐼2𝒅𝒍 ×𝒂 2 𝑹𝟐𝟏
4𝜋 𝑹𝟐𝟏2 ---(3)
Equation (3) represents the Coulombs law of force between the two current carrying elements.
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Force between two current carrying parallel conductors
Consider two straight parallel conductors of length l mts carrying current I1 and I2 respectively separated by a distance d mts as shown. The differential force dF1 exerted on
any differential element I1dl1 due the magnetic field B2 setup current I2 is given by 𝒅𝑭1 = 𝐼1𝒅𝒍1 × 𝑩2 = 𝐼1𝒅𝒍1𝑩2sin 𝜃 = 𝐼1𝒅𝒍1𝑩2sin 90𝑜 = 𝐼
1𝒅𝒍1𝑩2 The direction of force is perpendicular to both I and B, hence =90o
The total force 𝑭1 = 𝐼𝑙1 1𝑩2𝒅𝒍1= 𝐼1𝑩2𝒍1 (N)
for a long conductor, from Biot Savart law 𝑩𝟐= 𝝁 𝑰𝟐
𝟐𝝅𝒅- 𝑭 𝟏 = 0𝐿𝜇 𝐼12𝜋𝑑𝐼2𝒅𝒍1 =𝜇 𝐼2𝜋𝑑1𝐼2𝑙 = 𝑭 𝟐 Newtons
If I1 = I2 = I, then 𝑭 𝟏 = 𝜇𝐼2𝑙
2𝜋𝑑 = 𝑭 𝟐
For long conductors, force is expressed in N/mt as 𝑭 𝒍 = 𝜇 𝐼
2
2𝜋𝑑 N/mt
Magnetic Torque and Magnetic moment
The torque about a specified point is defined as the vector product of the moment arm 𝑹 and the force 𝑭 . It is measured in Newton meter.
𝑻 = 𝑹 × 𝑭 N-mt
Expression for Torque on a rectangular current loop
Consider a rectangular current loop of side dx and dy in x-y plane in an uniform magnetic field B
(i) Torque on side AB and side CD
dFAB = dF1 = Idl B = Idx ax [ Bxax + Byay + Bzaz ]
= Idx ( Byaz – Bzay)
The perpendicular distance for the side AB is 𝑹 1 = 𝑑𝑦 2 (− 𝑎 𝑦) The torque on side AB is dT1 = R1 dF1
dT1 = 𝑑𝑦
2 (− 𝒂 𝑦) Idx ( Byaz – Bzay) = − 𝐼 𝑑𝑥 𝑑𝑦
2 𝐵𝑦𝒂 𝑥 due to symmetry, dT1 = dT3 ( Torque on side CD)
(ii) Torque on side BC and side DA
dFBC = dF2 = Idl B = Idy ay [ Bxax + Byay + Bzaz ]
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The torque on side BC is dT2 = R2 dF2 dT2 =
𝑑𝑥
2 𝑎 𝑥 Idy ( Bz ax – Bx az) = 𝐼 𝑑𝑥 𝑑𝑦
2 𝐵𝑥𝑎 𝑦 due to symmetry, dT2 = dT4 ( Torque on side DA)
Total Torque dT = dT1 + dT2 + dT3 + dT4 = Idx dy [𝐵𝑥𝑎 𝑦 − 𝐵𝑦𝒂 𝑥]
dT = Idx dy az [ Bxax + Byay + Bzaz ] dT = I dsz B Nt-mt
We define magnetic dipole moment as m = I ds an,
hence dT = m B Total torque T = I S B Newton-mt
Review Questions
1. Derive Lorentz force equation and mention the application of the solution.
2. Derive an expression force on a differential current element placed in a magnetic field. 3. Derive an expression for the force between two differential current elements.
4. Deduce an expression for the force between two long conductors carrying the currents in same direction. Comment on the nature of force.
5. Derive an expression for the torque on a rectangular current loop is placed in a magnetic field.
Problems
1. The point charge Q=18nC has a velocity 5×106 m/s in the direction av=0.6ax+0.75ay+0.3az Calculate the
magnitude of the force exerted on the charge by the field (a) B=3ax – 4ay+6az mT (b) E=[3ax – 4ay+6az ] ×10 6
V/mt (c) When both B & E acting together.
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝒗 = 𝒗 𝒂 𝒗 = 5 × 106 0.6 𝒂 𝑥+ 0.75 𝒂 𝑦+ 0.3 𝒂 𝑧 = 3 𝒂 𝑥 + 3.75 𝒂 𝑦+ 1.5 𝒂 𝑧 × 106 𝑚𝑡/𝑠𝑒𝑐 = 297 𝒂 𝑥 − 405 𝒂 𝑦+ 418.5 𝒂 𝑧 µ𝑁 𝑭 𝑒 = 𝑄. 𝑬 = 18 × 10−9 −3 𝒂 𝑥+ 4 𝒂 𝑦 + 6 𝒂 𝑧 × 103= −54 𝒂 𝑥 + 72 𝒂 𝑦+ 108 𝒂 𝑧 µ𝑁 𝑭 = 𝑭 𝒆+ 𝑭 𝒎 = 243 𝒂 𝑥 − 333 𝒂 𝑦 + 526.5 𝒂 𝑧 µ𝑁 𝑭 = 2432+ (−333)2+ 526.52 = 669 µ𝑁 In general dT = I ds an B = I dsn B (N –mt) Or T = I S an B
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2. In a certain region of space B =𝟑𝒙 𝒂 𝒛 𝝁𝑻 , Find the force exerted on a
rectangular current loop in z=0 plane bounded by x =1 and x = 3; y =1 and y = 3 carrying a current of 2 mA.
Solution: 𝑭 = 𝐼𝒅𝒍𝒍 × 𝑩 = − 𝑩𝒍 × 𝐼𝒅𝒍
3. Find the force on a 4mt long conductor on z-axis with a current of 5A in az direction if B= 2ax+6ay (T)
Solution: 𝑭 = 𝐼𝒅𝒍 × 𝑩 = 5(4𝒂 𝑧) × 2𝒂 𝑥+ 6𝒂 𝑦 = −120𝒂 𝑥 + 40𝒂 𝑦 𝑁𝑒𝑤𝑡𝑜𝑛𝑠 𝑭 = −120𝟐+ 402 = 126.5 𝑵
Direction of force is given by 𝒂 𝐹 = −120𝒂 𝑥+40𝒂 𝑦
126.5 = −0.95 𝒂 𝑥+ 0.316 𝒂 𝑦
4. A conductor of length 2.5 mt located at x = 4, z = 0 carries a current of 12 A in – ay direction. Find the
value of B if the force on the conductor is 12mN in the direction – 0.707 ax + 0.707 az
Solution: 𝑭 = 𝐼𝑑𝑙 × 𝐵 = 12 2.5 −𝑎 𝑦 × (𝐵𝑥𝑎 𝑥+ 𝐵𝑦𝑎 𝑦 + 𝐵𝑧𝑎 𝑧) 12 × 10−3 (−0.707𝑎
𝑥 + 0.707𝑎 𝑧) = 30(𝐵𝑥𝑎 𝑧 − 𝐵𝑧𝑎 𝑥) Equating the ax and az components, we get Bx = Bz = 0.2828 mT B = 0.2828𝑎 𝑥+ 0.2828𝑎 𝑧 mT
5. Find the force per meter length between two parallel long conductors separated by 10cm in air carrying a current of 100 A in the same direction. Comment on the nature of the force.
Solution: 𝑭 𝒍 =
𝜇𝐼2
2𝜋𝑑 =
4𝜋×10−7(100)2
2𝜋(0.1) = 0.02 𝑁/𝑚𝑡 . Force is attractive in nature. 6. An electron with a velocity v = [3ax + 12ay – 4az ] ×10
5
mt/sec experiences a no force at point in the magnetic field B=[10ax + 20ay+ 30az ] mT. Find E at that point.
Solution:
F = Fe + Fm = Q (E+ V×B) = 0 E = B×V= = (–11.6 ax + 4.9 ay+ 6.0 az ) kV/mt
7. A conductor is located at x=0.4 mt 0<z<2 mt carries a current of 5 A in az direction in the field of B=2.5ax (T).
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Solution: F = Idl × B = Idz az × B = (5)(2) az ×2.5 ax = 25 ay N Torque on z axis T = R × F = 0.4 ax×25 ay = 10 az N-mt
8. A current loop in z =0 plane has a dimensions of (1×2) mt2 lies in a uniform magnetic
field B=[- 0.6 ay+ 0.8az ] T carries a current of 4mA. Find the torque exerted.
Solution: T = I Sn × B = 4 (2) az × (- 0.6 ay+ 0.8az ) m N-mt
T = 4.8 ax m N-mt
9.Find the maximum torque on a orbiting charged particle if the charge is 1.6×10-19 C, the circular path has a radius 0.5×10-10 mt, the angular velocity is 4×1016 rad/sec and B = 0.4mT.
Solution: The orbiting charge has a magnetic moment 𝒎 given by
𝒎 = {(2𝜋𝜔)𝑄} 𝑺 𝒂 𝒏=𝟒×𝟏𝟎𝟐𝝅𝟏𝟔 1.602 × 10−19 (𝜋 × 0.5 × 10−10 2 𝒂
𝒏 = 8.01 × 10−24 𝒂 𝒏 Amp-mt2 The maximum torque results when 𝒂 𝒏is normal to B.
Tmax = m B = 3.20× 10–27 N-mt
10. The coil is in YZ plane in the field of B= 2 (ax + ay) T, find the torque about z
axis.
Soln: the loop is in y-z plane. Area of the loop is S=ab ax
Torque T = N( I S × B) where N is the number of turns
T = N I ab ax × 2 (ax + ay) = 2 N I ab az N mt
11. For the square loop of wire z = 0 plane carrying 2 mA in the field of an infinite filament on y- axis as shown in the fig, Calculate the total force exerted on the loop.
Solution: The Magnetic flux density due to an infinite filament carrying a current of I amp in the plane of the loop is given by
𝐵 = 𝜇𝑜𝐼 2𝜋𝑥 𝑎 𝑧 = (4𝜋×10−7)15 2𝜋𝑥 𝒂 𝒛 = 3 𝑥 𝒂 𝑧 µ 𝑇
The total force on the loop is given by 𝑭 = 𝐼𝒅𝒍𝒍 × 𝑩
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Unit 5(b) Magnetization and Permeability
The electrons revolve in orbits around the +ve nucleus, they also rotate about their own axes. The movement of bound charges (orbital electrons, electron spin and nuclear spin) produces internal magnetic field similar to that produced by a current loop.
The magnetic moment of such a current loop is given by m = IS
The current produced by the bound charges is called bound current; the field produced due the movement of bound charges is called Magnetization represented by M. The Magnetization M is the dipole moment per unit volume as the volume shrinks to zero.
𝑀𝑎𝑔𝑛𝑒𝑡𝑖𝑠𝑎𝑡𝑖𝑜𝑛 𝑴 =𝑻𝒐𝒕𝒂𝒍 𝑴𝒂𝒈𝒏𝒆𝒕𝒊𝒄 𝒅𝒊𝒑𝒐𝒍𝒆 𝒎𝒐𝒎𝒆𝒏𝒕
𝑽𝒐𝒍𝒖𝒎𝒆 = 𝐥𝐢𝐦∆𝒗→𝟎
𝒎𝒊
∆𝑽 𝑨/𝒎𝒕
The magnetic flux density increases due to magnetization and for a general medium 𝑩 = 𝜇𝑜 [𝑯 + 𝑴 ] ---Note1
For a linear, isotropic and homogeneous material, M = m H where m is called Magnetic Susceptibility of the medium, which gives the measure of how sensitive the material to the magnetic field.
Hence 𝑩 = 𝜇𝑜 𝑯 + 𝑴 = 𝜇𝑜 𝑯 +𝑚 𝑯 = 𝜇𝑜 𝟏 +𝑚 𝑯 = 𝜇𝑜𝜇𝑟𝑯 = 𝜇𝑯 Where µr = (1 + m) is called relative permeability of the material in H/mt and 𝜇𝑟 =𝜇𝜇
𝑜
Classification of Magnetic materials
µr <1, m < 0 µr > 0, m > 0 µr >> 1, m >> 0
Ex: Lead, Copper, Sodium chloride Air, Tungsten, platinum Iron, Cobalt, Nickel & their alloys
Note1: Consider a diff. volume v in a material, in the absence of external field, the magnetic dipole moments are randomly oriented and net dipole moment is zero; hence M =0. With the application of the external field, the magnetic moments tend to align along the direction of the field; hence the net dipole moment is not zero. Consider the alignment of a magnetic dipole moment along a closed path dl as shown. The magnetic moment m makes an angle with the path. Let N be the total
number of dipoles along the path, then the total current strung along the path dl is given by dIb= Ib(Nds Cos ).dl = M.dl
The current over the entire contour 𝐼𝑏= 𝑴𝑐 . 𝒅𝒍
The Total current thro the material is given by
IT = Ifree charges + I bound charges
From Ampere’s law 𝜇𝑜 𝑯𝑐 . 𝒅𝒍 = 𝑩𝑐 . 𝒅𝒍 = 𝜇𝑜 𝑱 . 𝒅𝒔𝑠 + 𝑴𝑐 . 𝒅𝒍
(𝜇 𝑩
𝑜
𝑐 − 𝑴 ). 𝒅𝒍 = 𝑱 . 𝒅𝒔𝑠 = 𝐼 = 𝑯𝑐 . 𝒅𝒍 (from Ampere’s law for steady fields) Hence 𝑯 ≅ 𝑩
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The conditions of the field that exists at the boundary between the two media, when a magnetic field passes from one medium to another are called Magnetic Boundary conditions.
The vectors B & H at the boundary are resolved into tangential and Normal components. Consider the boundary between two media of permeability µ1 and µ2 as shown.
Normal Components
Imagine a Gaussian surface in the form of a right circular cylinder of height h so that half of the cylinder lies in the medium 1 and other half in medium 2.
From Gauss’s law 𝑩. 𝒅𝒔 = 𝟎 𝑜𝑟 𝑩. 𝒅𝒔𝑠 𝑡𝑜𝑝 + 𝑏𝑜𝑡𝑡𝑜𝑚 𝑩, 𝒅𝒔+ 𝑩. 𝒅𝒔𝑙𝑎𝑡
As Limit h0; 𝑩. 𝒅𝒔𝑙𝑎𝑡 = 0; hence Bn1.s – Bn2.s =0 Bn1 = Bn2
The normal components of the magnetic flux densities are continuous across the boundary
Hence 𝑯𝒏𝟏 𝑯𝒏𝟐 = 𝝁𝟐 𝝁𝟏 = 𝝁𝒓𝟐 𝝁𝒓𝟏 Tangential components
Applying ampere’s law over a closed path a-b-c-d-a
𝑯. 𝒅𝒍 = 𝑯. 𝒅𝒍𝑐 𝒂𝒃 + 𝑯. 𝒅𝒍𝒃𝒄 + 𝑯. 𝒅𝒍𝒄𝒅 + 𝑯. 𝒅𝒍 = 𝑰𝒅𝒂
As Limit h 0; The contributions from the vertical sides cd and da are zero. Hence 𝑯. 𝒅𝒍 = 𝑯. 𝒅𝒍𝑐 𝒂𝒃 + 𝑯. 𝒅𝒍𝒄𝒅 = 𝐻𝑡𝑎𝑛 1∆𝑤 − 𝐻𝑡𝑎𝑛 2∆𝑤 = 𝑲 ∆𝑤 𝐻𝑡𝑎𝑛 1− 𝐻𝑡𝑎𝑛 2 = 𝑲 𝐴𝑛𝑑 𝐵𝑡𝑎𝑛 1 𝜇1 − 𝐵𝑡𝑎𝑛 2 𝜇2 = 𝑲
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𝑩𝒕𝒂𝒏𝟏
𝝁𝟏 −
𝑩𝒕𝒂𝒏𝟐
𝝁𝟐 = 𝒂 𝒏𝟏𝟐× 𝑲
where 𝒂 𝒏𝟏𝟐 is a unit vector normal to the surface and is directed from medium1 to medium 2 If the media are not conductors, then K =0
𝐻𝑡𝑎𝑛 1− 𝐻𝑡𝑎𝑛 2 = 𝟎 𝐴𝑛𝑑 𝐵𝑡𝑎𝑛 1
𝐵𝑡𝑎𝑛 2 =
µ𝑟1
𝜇𝑟2
Tangential components of magnetic field intensities are continuous a/c a Dielectric –Dielectric boundary. 𝑡𝑎𝑛𝛼1 =𝐵𝑡𝑎𝑛 1 𝐵𝑛 1 𝑡𝑎𝑛𝛼2 = 𝐵𝑡𝑎𝑛 2 𝐵𝑛 2 or 𝑡𝑎𝑛 𝛼1 𝑡𝑎𝑛 𝛼2 = 𝐵𝑡𝑎𝑛 1 𝐵𝑡𝑎𝑛 2 = 𝜇𝑟1 𝜇𝑟2
Comparison between Electric and Magnetic circuits
Electric Circuits Magnetic Circuits
The path traced by the current is called electric circuit
The path traced by the magnetic flux constitutes a magnetic circuit
E= V;
Electro motive force (EMF) is the driving force measured in volts
H = Vm
Vm is called Magneto motive force (MMF) in magnetic circuits, expressed in Ampere Turns (AT) The line integral of Electric field over a path is
nothing but the EMF or Electric potential
𝑉𝑎𝑏 = 𝐸. 𝑑𝑙 𝑏
𝑎
The line integral of magnetic field intensity over a path is called MMF 𝑉(𝑚 −𝑎𝑏 )= 𝐻. 𝑑𝑙 𝑏 𝑎 = 𝑚𝑚𝑓 𝑱 = 𝜎 𝑬 ---- Ohm’s law B = µ H Electric current 𝐼 = 𝑱. 𝒅𝒔𝑠 𝑀𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑓𝑙𝑢𝑥 ∅ = 𝑩. 𝒅𝒔 𝑠 Resistance 𝑅 =𝑉𝐼 Ohms ; 𝑅 =𝜎𝑆𝑙 Reluctance ℜ =𝑉𝑚
∅ AT/web; ℜ = 𝑙 𝜇𝑆 𝐸. 𝑑𝑙 = 0 𝑙 𝐻. 𝑑𝑙 = 𝐼𝑙 𝑒𝑛𝑐𝑙 = 𝑁𝐼 (for N turns)
Differences between Electric and Magnetic circuits:
1. In electric circuits, the current actually flows due to the movement of electrons. While in magnetic circuits, due to MMF the flux gets established and does not flow.
2. Energy must be supplied to the electric circuit to maintain the flow of current, where as in magnetic circuits, energy must be required to create the flux, but not required to maintain it.
3. Electric flux lines not closed and an isolated charge can exist, where as magnetic flux lines are always closed and isolated magnetic pole cannot exist.
4. Resistance and conductivity varies with temperature, where as reluctance and permeance are independent of temperature variations.
5. Permeability µ is dependent on the flux density B, where as the conductivity is independent of current density J.
Difficulties in analysing magnetic circuits
Magnetic circuits are generally made up of ferromagnetic materials and relative permeability of these materials depends on flux density.(non linearity). It is difficult to control and calculate the leakage flux. If there is an air gap in the magnetic path, the flux spreads and bulges out leading to fringing effect.
This is a supporting material for the SCE students. Any use for commercial purpose has to be permitted by the author ©Ravi Shankara.M.N. Page 11 Magnetic Energy and Energy density
Consider a diff. volume V = xyz mt3 in a magnetic field B as shown.
Let the top and bottom surfaces caries a current of I amps. Inductance of the diff. volume is defined as
𝐿 = ∆𝐼 = 𝐵.∆𝑠∆𝐼 = 𝐵.∆𝑥∆𝑧∆𝐼 = 𝜇𝐻∆𝑥∆𝑧∆𝐼
From ampere’s law, I = H.y ( 𝐻. 𝑑𝑙 = 𝐼 𝑖𝑛 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑓𝑜𝑟𝑚)𝑙 We know that Energy stored in the inductor L is given by Wm = 12 ∆𝐿 ∆𝐼2
∆𝑊𝑚 = 1 2 𝜇𝐻∆𝑥∆𝑧 ∆𝐼 ∆𝐼2 = 1 2𝜇𝐻∆𝑥∆𝑧 𝐻∆𝑦 = 1 2𝜇𝐻2∆𝑉 Energy density ∆𝑊∆𝑉𝑚 = 12𝜇𝐻2= 12𝐵 2 𝜇 = 1 2𝐵𝐻 Joules/mt 3
Energy stored in the magnetic field Wm= ∆𝑊𝑚𝑑𝑣 =
1 2 𝜇𝐻2𝑑𝑣 = 1 2 𝐵𝐻𝑣𝑜𝑙 𝑣𝑜𝑙 𝑑𝑣 Joules II method
The voltage a/c the inductor is given by 𝑉 = 𝐿𝑑𝑡𝑑𝑖
and the current through the inductor is given by 𝑖 = 1𝐿 𝑉 𝑑𝑡
We know that inductor stores the energy in magnetic field, and is given by 𝑊 = 𝑃. 𝑑𝑡 = 𝑉. 𝑖 𝑑𝑡 = 𝐿 𝑖 𝑑𝑖0I = 12𝐿𝐼2 Joules
Also we define inductance as 𝐿 = 𝑐𝑢𝑟𝑟𝑒𝑛𝑡𝐹𝑙𝑢𝑥 = ∅𝐼 = 𝐵.𝐴𝐼
Also from Ampere’s law, H.l = I, hence we can rewrite the magnetic energy stored as 𝑊 = 12𝑩 𝐴 𝑯 𝑙 =12 𝑩 𝑯 𝐴 𝑙 = 12 B H (Vol)
In differential form 𝑑𝑤 = 12𝑩𝑯𝑑𝑣 and in integral form 𝑊 = 12 𝑩. 𝑯 𝑑𝑣 Joules The term energy per unit volume is known as Energy density expressed in J/m3
𝐸𝑛𝑒𝑟𝑔𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑊𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =12𝑩. 𝑯 =12𝜇𝐻2 = 2𝜇1 𝑩2 J/m3
Forces on magnetic materials Consider an electromagnet with N turns carrying a steady current I as shown. Due to magnetic boundary conditions, the flux in the air gap is same as that in the core. Bn1 =Bn2.The two circuits are separated dl by distance as shown. The energy required to separating the two circuits by a distance
This is a supporting material for the SCE students. Any use for commercial purpose has to be permitted by the author ©Ravi Shankara.M.N. Page 12 −𝐹. 𝑑𝑙 = 12𝐵𝜇2 𝑜𝑑𝑣 2 = 1 2 𝐵2 𝜇𝑜𝑨. 𝑑𝑙 2 = − 𝐵2 𝜇𝑜𝑨 𝑁𝑒𝑤𝑡𝑜𝑛𝑠
Where A is the Cross sectional area of the gap and factor of 2 used to account for two air gaps. The –ve sign indicates that the force acts to reduce the air gap.
Hence force on the single air gap is 𝑭 =12 𝐵𝜇2
𝑜𝐴
In electromechanical systems, the force is expressed in terms of pressure as 𝑭 𝑨= |𝑷| = 1 2 𝐵2 𝜇𝑜 = 1 2𝐵𝐻 = 1 2𝜇𝑜𝐻2 N/mt2 Inductance and mutual Inductance
Inductors are the conductors which stores energy in the magnetic field. Inductance of an inductor is the opposition offered by an inductor for the flow of rate of change of current through it. Inductance is expressed in Henries.
When current flows in an inductor, the flux lines links the Inductor itself, hence the inductance is called as self inductance.
Inductance of an inductor is defined as the ratio of total flux linkages to the current producing it. 𝑰𝒏𝒅𝒖𝒄𝒕𝒂𝒏𝒄𝒆 𝑳 =𝝀𝑰= 𝑵∅𝑰 Henries
The Voltage-Current relation in an inductor is given by 𝑣 = 𝐿𝑑𝑡𝑑𝑖
Power P = v i = 𝐿𝑑𝑖𝑑𝑡 𝑖
Energy stored in the inductor 𝑊𝑚 = 𝑃. 𝑑𝑡 = 𝐿 𝑖 𝑑𝑖𝑖=0𝐼 =12 𝐿𝐼2 Joules
Inductance of an inductor depends upon the length of the inductor, Number of turns, the permeability, Area of cross section.
Inductance of the Solenoid
Consider a long solenoid of N turns, having cross sectional area A carrying a current of I amp.
The magnetic field at any point on the axis of the long solenoid is given by 𝑩 = 𝝁𝑵𝑰𝒍 (T)
The total flux λ = N = N B A = 𝝁𝑵𝟐𝒍𝑰 𝑨
𝐼𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡𝑒 𝑆𝑜𝑙𝑒𝑛𝑜𝑖𝑑 𝑳 = 𝑻𝒐𝒕𝒂𝒍 𝒇𝒍𝒖𝒙𝑪𝒖𝒓𝒓𝒆𝒏𝒕 =𝝁𝑵𝒍𝟐𝑨 𝐻𝑒𝑛𝑟𝑖𝑒𝑠
Inductance of a Toroid
Consider a Toroidal coil having N turns carrying current I amp, having circular cross sectional area A as shown in fig. The magnetic field at the centre of the Toroid is given by 𝑩 = 𝜇𝑁𝐼2𝜋𝑅 wb/mt2
This is a supporting material for the SCE students. Any use for commercial purpose has to be permitted by the author ©Ravi Shankara.M.N. Page 13 The flux linkages = λ = N = N B A = 𝜇 𝐴 𝑁
2𝐼
2𝜋𝑅 wb Inductance of the Toroid 𝐿 = 𝜆𝐼 = 𝑁𝜑𝐼 = 𝜇 𝑁2𝜋𝑅2𝐴
Note1: For a circular cross section, A = r2,
𝐿 = 𝜇 𝑁2𝜋𝑅2𝐴 =𝜇 𝑁2𝑅2 𝑟2 where R is the mean radius
Note2: For a Toroid with N turns, h as the height, with r1 as inner radius and r2 as outer radius, Square cross section 𝐿 = 𝜇 𝑁2𝜋2 ln (𝑟2
𝑟1) Henries
Inductance of a Coaxial cable
Consider a Coaxial cable along z axis with co-axial conducting cylinders of radii a and b as shown, carrying a steady current of I amp
The Flux density in between the cylinders is given by (for the region a < r < b) 𝑩 = 2𝜋𝑟𝜇𝐼 𝒂 𝝋 ;
The total flux in the region is given by 𝝋 = 𝑩. 𝒅𝒔 =𝒔 𝟐𝝅𝝁𝑰 𝒂𝒃𝒅𝒓𝒓 𝑑𝑧𝟎𝒅
𝜑 =𝜇𝐼𝑑2𝜋 ln 𝑏𝑎 𝑤𝑏
Inductance of the Coaxial cable
𝐿 = 𝐿𝑒𝑥𝑡 = 𝑇𝑜𝑡𝑎𝑙 𝑓𝑙𝑢𝑥 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = 𝜇𝑑 2𝜋ln 𝑏 𝑎 𝐻𝑒𝑛𝑟𝑖𝑒𝑠 Considering the internal inductance
The flux density at any point inside (r < a) the solid inner conductor of radius a is given by 𝑩 =2𝜋𝑎𝜇𝐼𝑟2𝒂 𝝋
Energy stored in the Inductor W = 12𝐿𝑖𝑛𝑡𝐼2 𝐿𝑖𝑛𝑡 = 𝐼22 𝑊 =𝑰𝟐𝟐 𝑩
𝟐 𝟐𝝁 𝒗𝒐𝒍 𝑑𝑣 𝑳𝒊𝒏𝒕 =𝑰𝟏𝟐 𝟎𝒂 2𝜋𝑎𝜇𝐼𝑟2 𝟐𝑟 𝑑𝑟 𝑑𝜑 𝑑𝑧 =02𝜋 0𝑑 𝜇𝑑8𝜋 𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝐿 = 𝐿𝑖𝑛𝑡 + 𝐿𝑒𝑥𝑡 = 𝜇𝑑 8𝜋+ 𝜇𝑑 2𝜋ln 𝑏 𝑎 = 𝜇𝑑 2𝜋 1 4+ ln 𝑏 𝑎 Mutual Inductance
Consider two circuits 1 and 2 having turns N1 and N2 carrying currents I1 and I2 respectively. 11 is the flux linking with circuit1 due current I1, 22 is the flux linking with circuit 2 due current I2 ,12 is the flux linking with circuit 2 due current I1, 21 is the flux linking with circuit1 due current I1 The mutual inductance between two coils with respect to flux linkages is defined as the ratio of flux linkage of one coil to the current in the other coil.
This is a supporting material for the SCE students. Any use for commercial purpose has to be permitted by the author ©Ravi Shankara.M.N. Page 14 𝑀12 = 𝑁2𝐼𝜑12
1 (𝐻) 𝑀21 =
𝑁1𝜑21
𝐼2 (𝐻)
Where N212 – Flux linkage in ckt 2 due to current I1 in the ckt1. N121 – Flux linkage in first circuit due to current I2 in the second ckt.
Expression for Mutual Inductance between two concentric solenoids
Consider two concentric long solenoids of length l1 and l2 mts with radii r1 and r2, turns N1 and N2 respectively carrying a steady current I1 and I2 amps.
𝑩 𝑠𝑜𝑙𝑒𝑛𝑜𝑖𝑑 1 = 𝑩𝟏 = 𝜇𝑁𝑙1𝐼1
1
Flux linking with the solenoid2 due to I1 = 𝑁2𝜑12 = 𝑁2𝐵1𝐴1 =
𝜇 𝑁1𝑁2𝐼1 𝑙1 𝜋𝑟1 2 Mutual inductance 𝑀12 = 𝑁2𝜑12 𝐼1 = 𝜇 𝑁1𝑁2 𝑙1 𝜋𝑟1 2 similarly 𝑀 21 = 𝑁1𝐼𝜑21 2 = 𝜇 𝑁1𝑁2 𝑙2 𝜋𝑟2 2 Review Questions
1. Define (i) Magnetization (ii) Permeability. 2. Compare electric and magnetic circuits.
3. Derive the magnetic boundary conditions at the interface between two different magnetic materials. Discuss the conditions.
4. Define self inductance and mutual inductance with suitable formulae. 5. Derive an expression for the self inductance of a long Solenoid. 6. Derive an expression for the self inductance of a Toroid.
7. Derive an expression for the self inductance of a co-axial cable.
Note
This is a supporting material for the SCE students. Any use for commercial purpose has to be permitted by the author ©Ravi Shankara.M.N. Page 15 Problems on Magnetisation, permeability, boundary conditions
1. Given a material for which χm = 3.1 and within which B = 0.4y az T, find: (a) H (b) μ (c) μr (d) M (e) J
Soln: 𝑎 𝐵 = 𝜇𝑜 1 + 𝜒𝑚 𝐻 → 𝐻 = 𝜇 𝐵 𝑜 1+𝜒𝑚 = 0.4 𝑦 𝑎𝑧 𝜇𝑜 1+3.1 = 77.65 𝑦 𝑎𝑧 𝑘𝐴/𝑚𝑡 𝑏 𝜇 = 𝜇𝑜 1 + 𝜒𝑚 = 𝜇𝑜 1 + 3.1 = 51.5 µ𝐻/𝑚𝑡 𝑐 𝜇𝑟 = 1 + 𝜒𝑚 = 4.1 (d) M = 𝜒𝑚 𝐻 =3.1(77.65 𝑦 𝑎𝑧) = 241 𝑦 𝑎𝑧 𝑘𝐴/𝑚𝑡 (e) J=×H = 𝜕𝐻𝜕𝑦𝑧𝑎𝑥 = 77.65 𝑎𝑧 𝑘𝐴/𝑚𝑡2
2. Find the magnetic field intensity inside the magnetic material for the following conditions (a) M= 100 A/mt, µ= 15µH/mt (b) B= 300µT, m=15 (c) There are 8.28 ×1020 atoms/mt3, each atom has a dipole
moment of 5 × 10-27 A mt2 and µr =30. Soln: 𝝁𝒓 = 𝝁 𝝁𝒐 = 𝟏𝟓×𝟏𝟎−𝟔 𝟒𝝅×𝟏𝟎−𝟕 = 𝟏𝟏. 𝟗𝟒 𝝁𝒓 = 𝟏 + 𝜒𝒎 𝜒𝒎 = 𝝁𝒓− 𝟏 = 𝟏𝟎. 𝟗𝟒 (a) 𝑯 =𝜒𝑴 𝒎 = 𝟏𝟎𝟎 𝟏𝟎.𝟗𝟒= 𝟗. 𝟏𝟒 𝑨/𝒎𝒕 (b) Given B= 300µT, m=15 𝝁𝒓 = 𝟏 + 𝜒𝒎 = 𝟏𝟔 𝑯 =𝝁𝑩 𝒐𝝁𝒓= 𝟑𝟎𝟎×𝟏𝟎−𝟔 𝟒𝝅×𝟏𝟎−𝟕×𝟏𝟔= 𝟏𝟒. 𝟗𝟑 𝑨/𝒎𝒕 (c) M =mN =5×10-27×8.28×1020 = 4.14×10-6 A/mt 𝑯 =𝜒𝑴 𝒎 = 𝑴 𝝁𝒓−𝟏= 𝟒.𝟏𝟒×𝟏𝟎−𝟔 𝟑𝟎−𝟏 = 𝟎. 𝟏𝟒𝟐 × 𝟏𝟎−𝟔 𝑨/𝒎𝒕
3. Find the magnitude of magnetization in a material for which: (a) the magnetic flux density is 0.02 Wb/m2 and the magnetic susceptibility is 0.003 (b) the magnetic field intensity is 1200 A/m and relative permeability 𝝁𝒓 is 1.005. Soln: 𝑎 𝑀 = 𝜒𝒎 𝑩 𝝁𝒐𝝁𝒓= 𝜒𝒎 𝑩 𝝁𝒐(𝜒𝒎+𝟏)= 𝟎.𝟎𝟐 𝝁𝒐 (𝟎.𝟎𝟎𝟑) (𝟏+𝟎.𝟎𝟎𝟑)= 𝟒𝟕. 𝟔 𝑨/𝒎𝒕 (b) 𝑴 = 𝑯. 𝜒𝒎= 𝑯. (𝜇𝒓− 𝟏) = 𝟏𝟐𝟎𝟎 𝟏. 𝟎𝟎𝟓 − 𝟏 = 𝟔 𝑨/𝒎𝒕
4. The region z < 0 is characterised by B = 150 ax – 400 ay + 250 az; µr1= 5 and for the region z > 0, µr2= 2.
Determine the values of tangential and normal components of the field and their corresponding angles with respect to normal in both the regions.
Soln: Since z = 0 is the boundary, normal components are along az direction and also K=0 (Assumed)
Given B = B1 = 150 ax – 400 ay + 250 az
Bn1= 250 az and Btan1 = 150 ax – 400 ay | Btan1| = 427.2
Using the boundary relations 𝑩B𝐭𝐚𝐧𝟏
𝒕𝒂𝒏𝟐 = 𝝁𝒓𝟏 𝝁𝒓𝟐 ==> 𝑩𝒕𝒂𝒏𝟐 = B𝐭𝐚𝐧𝟏 𝝁𝒓𝟐 𝝁𝒓𝟏= [150 𝐚x – 400 𝐚y]× 0.4 𝑩𝒕𝒂𝒏𝟐 = [60 𝐚x – 160 𝐚y] |𝑩𝒕𝒂𝒏𝟐| = 𝟏𝟕𝟎. 𝟗 and Bn1= 250 az = Bn2 B2 = 60 ax – 160 ay + 250 az 𝛼1 = 𝑡𝑎𝑛−1 B𝐵tan 1 𝑛 1 = 𝑡𝑎𝑛 −1 427.2 250 = 59.66 𝑜 𝛼 2 = 𝑡𝑎𝑛−1 B𝐵tan 2 𝑛 2 = 𝑡𝑎𝑛 −1 170.9 250 = 34.35 𝑜
This is a supporting material for the SCE students. Any use for commercial purpose has to be permitted by the author ©Ravi Shankara.M.N. Page 16 5. The z = 0 plane marks the boundary between two magnetic materials. Region 1 is defined by z > 0 and the magnetic field intensity in this region is 40 ax +50 ay + 12 az kA/mt. Region 2 is defined by z < 0 and
has a relative permeability of 1000. If the relative permeability of the medium is 200, find the field intensity in medium 2. A current sheet of 12 ay kA/mt is present at the boundary.
Solution: z = 0 is the boundary; normal components are along az direction.
Normal component of K is along (-az) direction
H tan2 = (40 ax + 50 ay) kA/mt and Hn2 = 12 az kA/mt
From the magnetic boundary conditions
H tan1 - H tan2 = K = an12 × K= ( -az )× [12 × 103 ay]= 12 ax kA/mt
Htan1 = Htan2 – 12 ax= (40 ax + 50 ay) – 12 ax = (28 ax + 50 ay) kA/mt
Htan1 = (28 ax + 50 ay) kA/mt 𝑯𝒏𝟏 𝑯𝒏𝟐= 𝝁𝒓𝟐 𝝁𝒓𝟏 𝑯𝒏𝟏 = 𝝁𝒓𝟐 𝝁𝒓𝟏𝑯𝒏𝟐= 200 1000 (12𝐚z)= 2.4 𝒂𝒛 kA/mt H1 = 28 ax +50 ay + 2.4 az kA/mt
6. Let the permeability be 5 µH/mt in the region A where x < 0 and 20 µH in the
region B where x > 0. If there is a surface current density K = 150 ay – 200 az A/mt at x=0. Find the
normal and tangential components of H in both the regions if HA = 300 ax – 400 ay + 500 az,
Soln: Since x = 0 is the boundary, normal components are along ax direction.
Normal component of K is along (ax) direction as
H tanA = – 400 ay +500 az and HnA = 300 ax From the magnetic boundary conditions
HtanA – HtanB = K = an × K= ( ax )× [150 ay – 200 az]= 150 az + 200 ay
HtanB = HtanA + [150 az +200 ay] = – 400 ay +500az – 150 az – 200 ay
HtanB = – 600 ay + 350 az 𝑯𝒏𝑩 𝑯𝒏𝑨= 𝝁𝒓𝑨 𝝁𝒓𝑩 = 𝟓 𝟐𝟎 𝐻𝑛𝐵 = 𝝁𝒓𝑨 𝝁𝒓𝑩𝐻𝑛𝐴 = 5 20 300 𝐚x= 75 𝒂𝒙
Problems on Inductance and mutual inductance
1. Find the mutual inductance between a conducting loop and a very long straight conductor as shown in figure.
Solution: The magnetic field intensity at any point (distant r from the conductor) due to a long current carrying conductor is given by Ampere’s law 𝐻 𝑙 1. 𝑑𝑙 = 𝐼1 or 𝐻 1 = 𝐼1 𝑑𝑙𝑙 = 𝐼1 𝑟 𝑑∅ 2𝜋 0 𝒂 ∅ = 𝐼1 2𝜋𝑟𝒂 ∅ 𝐴/𝑚𝑡 The magnetic flux density 𝐵 1 = µI1
2πr 𝒂 ∅ 𝑇
This is a supporting material for the SCE students. Any use for commercial purpose has to be permitted by the author ©Ravi Shankara.M.N. Page 17 ∅12 = 𝐵 . 𝑑𝑠𝑠 = 𝑧=0𝑏 𝑟=𝑑𝑑+𝑎2πrµI1 𝒂 ∅. 𝑑𝑟 𝑑𝑧 𝒂 ∅ =µI2π1b 𝑟=𝑑𝑑+𝑎drr =µI2π1bln d + a − ln 𝑑 =µI2π1b 1 +da 𝑤𝑏 Mutual inductance 𝑀12 =∅I12 1 = µb 2π 1 + a d 𝐻𝑒𝑛𝑟𝑖𝑒𝑠
2. A toroidal core has a rectangular cross section defined by the surfaces r = 2 cm, r = 3 cm, z = 4 cm, and
z = 4.5 cm. The core material has a relative permeability of 80. If the core is wound with a coil containing
8000 turns of wire, find its inductance.
Soln: Applying Ampere’s law to a circular loop of radius r mts, 𝑓𝑙𝑢𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑩 = 𝜇𝑯 = 𝑁2𝜋𝑟𝜇𝐼 𝒂∅ Total flux linkages = N = = 𝑁 𝑩. 𝒅𝒔 = 𝑁 𝑁2𝜋𝑟𝜇𝐼 𝑑𝑟 𝑑𝑧𝑧𝑧2
1 𝑟2 𝑟1 𝑠 𝐼𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝐿 =𝑁∅𝐼 =𝜇𝑁2 𝑧2−𝑧1 2𝜋 ln 𝑟2 𝑟1 = 4𝜋×10−7 × 80002 0.045−0.04 2𝜋 ln ( 0.03 0.02) = 2.08 𝐻
3. A 200 turns toroidal coil with µr= 1000, has a mean radius of 20 cm with a circular cross section having
radius 1 cm. Find the desired current to produce a flux of 0.5 mwb in the core. Soln: AmpereTurns AT =NI = 𝕽 = ∅ 𝜇𝐴𝑙 = 0.5×10
−3 2𝜋×2×10−2
4𝜋×10−7×1000×𝜋×12×10−4 = 1592 AT
I =1592𝑁 = 1592200 = 7.96 𝐴
4. Determine the energy stored in the region between the two conductors in a coaxial cable. Soln: The magnetic field intensity in between the conductors (a < r < b) is given by 𝑯 =2𝜋𝑟𝐼 a A/mt Energy stored 𝑊 = 12𝜇𝐻2𝑑𝑣 = 𝑣𝑜𝑙 𝜇 2 [ 𝐼 2𝜋𝑟 𝑣𝑜𝑙 ]2 𝑟𝑑𝑟𝑑∅𝑑𝑧 = 𝜇 𝐼2 8𝜋2 𝑑𝑟 𝑟 𝑏 𝑟=𝑎 𝑑∅ 𝑑𝑧 𝑑 𝑧=0 2𝜋 0 = 𝜇𝑑 𝐼2 4𝜋 ln 𝑏 𝑎 𝐽 Energy stored per unit length = 𝑊𝑑 =𝜇 𝐼4𝜋2ln 𝑏𝑎 J/mt
5. Calculate the inductance of a solenoid of 200 turns wound tightly on a cylindrical tube of length 60 cm and diameter 6 cm.
Soln: 𝐿 =𝜇𝑜𝑁2𝐴𝑙 = 4𝜋×10
−7× 2002× 𝜋× 32×10−4
0.6 = 0.2368 𝑚𝐻
6. An air cored toroid has a circular cross section with radius 4 mm, having 2500 turns with a mean radius of 20 mm. find the inductance.
Soln: 𝐿𝑡𝑜𝑟𝑜𝑖𝑑 = 𝜇 𝑁 2𝐴 2𝜋𝑅 = 𝜇 𝑁2 𝜋𝑟2 2𝜋𝑅 = 𝜇 𝑁2𝑟2 2𝑅 = 4𝜋×10−7×25002×0.0042 2×0.02 = 3.14𝑚𝐻
7. An air cored toroid has 700 turns with inner radius 1 cm and outer radius 2 cm, height 1.5 cm. Find the inductance of the Toroid assuming air medium and cross section as square. Compare the result with the general formula of inductance of Toroid
Solution: For a square cross section Toroid, 𝐿 = 𝜇𝑜𝑁
2 2𝜋 ln 𝑟2 𝑟1 = 4𝜋×10−7(700)21.5×10−2 2𝜋 ln 2 1 = 1.04 𝑚𝐻 for a square cross section Toroid 𝑅 = 𝑟1+𝑟2
2 =
1+2
2 = 1.5 𝑐𝑚 and 𝐴 = 2𝜋𝑟 = 2𝜋 × 10−2 1.5 × 10−2 Inductance of a Toroid is given by , 𝐿 = 𝜇𝑜𝑁2𝐴
2𝜋 𝑅 =
4𝜋×10−7 700 2(2𝜋×10−2)1.5×10−2
This is a supporting material for the SCE students. Any use for commercial purpose has to be permitted by the author ©Ravi Shankara.M.N. Page 18 8. A Solenoid with N1= 1000, r1=1 cm and l1=50cm is concentric within a second coil of N2=2000, r2=2cm
and l2=50cm. Find the mutual inductance assuming air medium.
Soln: 𝑀12 = 𝑁2𝐼𝜑12
1 =
4𝜋×10−7×1000×2000
