Electromagnetism is a pretty advanced topic in physics. It lays the foundation of more advanced optics and quantum mechanics. It is therefore essential for a student to be familiar to the solved and unsolved questions so that they are able to approach the topic better.
Solved problems in Electromagnetics make use of various predefined laws and theorems. One of the very first is the Coulombs Law for charged particles. Gradually we move into Gauss' Law, Electric Field, Magnetic Force on a Charged Body, Faraday's Electromagnetic Induction etc. It will not be possible to cover the problems based on all the topics here, but we shall show some examples of the problems and their solutions.
Problem: A charge of Q1=-1.0μC is placed at the origin of the rectangular coordinate system and a second charge, Q2=-10mC is placed on the x-axis at a distance of 50 cm from thee origin. Find the force on Q1 due to Q2, if they are in free space.
Solution: Q1 =-1.0μC at (0, 0, 0) and Q2 =-10mC at (0.5, 0, 0)
By Coulombs law,
F12 = Q1Q2a12/4`pi` `epsi` r2 Where F12 is force between two charges,
Q1,Q2 are two charges,
r is distance between two charges,
a12 is unit vector along the line joining two charges.
r =(0, 0, 0)-(0.5, 0, 0) = -0.5ax
¦r¦ = 0.5
F12 = ( -1.0 * 10 -6 ) ( -10 * 10-3) 9 * 109 ( -ax)
F12 = - 360 ax Newton ( Answer )
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Problem: Two point charges Q1= 5.0C Q2= 1.0nC are located at (-1, 1, -3) m and (3, 1, 0) m, respectively. Determine the electric field at Q1.
Solution: Q1= 5.0C is at (-1, 1, -3) m
Q2=1.0nC is at (3, 1, 0) m
Electric field, E at Q1 (-1, 1, -3) m is
E = (Q2/4`pi` εr2) `xx` ar
r = (-1-3)ax +(1-1)ay +(-3-0)az
= -4ax -3az
r =¦r¦ =`sqrt(16+9)` =5
ar = (-4ax -3az) / 5
E = [1.0* 10-9 * 9 *109 / 25] `xx` (-4ax -3az) / 5
E = -0.288ax -0.216az V/m ( Answer ).
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Problem: An infinite sheet in XY plane extending from -`oo` to `oo` in both direction has a uniform charge density of 10 nC/m2. Find the electric field at z = 1.0 cm.
Solution: Surface charge density `rho` s = 10nC/m2.
For a sheet of charge lying in the XY plane, the field at any point on Z-axis is given by
E = `rho` s/2ε0 `xx` az
= (10*10-9/2*8.854*10-12) `xx` az
= 0.5647 * 103 az
E = 564 az V/m. (Answer )
Problem: 30 coulomb charge flows through a wire for 9 minutes. Calculate the current through the wire.
Solution: We know that current I = Rate of flow of charge = charge/time = Q/t
Q =30 C,t =9 minutes =9*60 seconds
Substituting values in the above relation
I =30/9*60 =0.05 ampere
0.05 ampere. (Answer )
Problem: 1350 joules work is done in circulating 3 amperes current for 5 minutes in an electric circuit. Obtain the e.m.f. of the source connected to the circuit.
Solution: Charge = Current * time
E.M.F E = Workdone/charge
I =3 amperes, t =5 minutes =5*60 seconds and W =1350 joules
Charge = 3*5*60 coulombs
E = 1350/3*5*60 = 1.5 Volts (Answer)
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Problem: A circular coil of radius 2.0 cm is in a magnetic flux density of 10 wb/m2. If the plane of a coil is perpendicular to the field, determine the total flux around the coil.
Solution: The flux density = 10 wb/m2
Area of the coil, S = `pi` r2
Total flux `phi` = B.S Where B =Flux Density and S =surface area
As B and S are in same direction,
B.S = BS
`phi` = 10 * 4`pi` *10-4
`phi` = 12.56 mwb (Answer)
Problem: If a wave with a frequency of 100 Mhz propagates in free space, find the propagation constant.
Solution: Propagation constant , `gamma` =`alpha` +j`beta` Where `gamma` is propagation constant
`alpha` is attenuation constant
`beta` is shift constant
For free space attenuation constant `alpha` is 0, therefore propagation constant `gamma` =j`beta`
Also `beta` =`omega` (μ0ε0)1/2 `omega` is angular frequency
μo is relative permittivity of free space
εo is relative permeability of free space