Mid term - III
1. In the figure, a beam of light, with intensity 35 W/m2 and polarization parallel to a y axis,
is sent into a system of two polarizing sheets with polarizing directions at angles of θ1 =
660and θ2 = 900 to the y axis. What is the intensity (in Watts/m2) of the light transmitted
by the two-sheet system?
Intensity through a polarizer at an angle θ is given by
I=I0cos2θ
So after the first sheet the intensity will be
I1= I0cos2θ1
Now the beam will be polarized with an angle θ1 with initial axis. And with angle
900- θ1 to the next polarizer. So the final intensity will be
If= I0 cos2θ1 cos2(900-θ1) => I0 cos2θ1 sin2θ1 = 4.8323 W/m2
2. Light with an intensity of 1kW/m2 falls normally on a surface with an area of 1 cm2 and
is completely reflected. The force of the radiation on the surface is:
Since its completely reflected, so force is F= (p-(-p))/∆t=2p/∆t
= 2 I A/c = 6.7 x 10-10
3. An arrangement for generating a traveling electromagnetic wave in the shortwave radio
region of the spectrum works as follows: an LC oscillator produces a sinusoidal current in
the antenna, which generates the wave, traveling outward at the speed of light. What is
the wavelength (in meters) of the wave emitted by this system if L = 0.323µH and
C = 45.0pF?
Wavelength λ = c/f = (2π c) /ω = (2π c) (LC)1/2 = 7.1864 m
4. A 1-µF capacitor is connected to an emf that is increasing uniformly with time at a rate of
100 V/s. The displacement current between the plates is:
Id= є0 (dΦE/dt) = є0 A/d (dV/dt) = C (dV/dt) = 1 x 10-4
5. A current of 1A is used to charge a parallel plate capacitor with square plates. If the area
of each plate is 0.6m2 the displacement current through a 0.3m2 area wholly between the
capacitor plates and parallel to them is:
I = J A = 0.5 A
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CHAPTER 33
6.
7. A charged capacitor and an inductor are connected in series. At time t = 0 the current is
zero, but the capacitor is charged. If T is the period of the resulting oscillations, the next
time after t = 0 that the energy stored in the magnetic field of the inductor is a maximum
is:
Q=Qmcos(2πt/T)
I=Imsin(2πt/T)
U=LI2/2 Umax=LI2m/2
For this to happen sine should be 1. That happens when
2πt/T = π/2 t=T/4
8. An LC circuit has an inductance of 15mH and a capacitance of 10µF. At one instant the
charge on the capacitor is 25µC. At that instant the current is changing at the rate of:
dI/dt = Q/LC = 166.6667 A/s
( 170 is closest)
9. An RLC series circuit has R = 4Ω, XC = 3 Ω, and XL = 6 Ω. The impedance of this circuit
is:
Z=
R 2 + ( X L − X C )2 = 5 Ω
10. Suppose this page is perpendicular to a uniform magnetic field and the magnetic flux
through it is 5Wb. If the page is turned by 30 degrees around an edge, the flux through it
will be:
ΦB= B A cosθ = 5 Wb cos(300) = 4.3 Wb
11. A long straight wire is in the plane of a rectangular conducting loop. The straight wire
carries a constant current i, as shown. While the wire is being moved toward the
rectangle the current in the rectangle is
Since the magnetic field due to the current is going inward the page, So the flux is
also going inward (-) . Towards the wire the magnetic field increases as 1/r so the
magnitude of the downward (negative) flux is increasing in magnitude, so from Lenz’s
law we know that the Induced magnetic field should oppose this change, that means
we need a Magnetic field that goes outward the page and this can be generated by only
an anticlockwise current.
An alternate mathematical approach : Lets take at a given time t closest edge is at a
distance x from wire. So flux through an infinitesimal element
1313
dφ = −
µ 0i
dx
Ldx = a
2πx
x
x +W
φ=
∫
a
x
dx
w+ x
= a ln(
)
x
x
 W
 dx
dφ
= a 
 > 0
dt
 (W + x ) x  dt
Since the emf is positive so the induced current is counter-clockwise.
emf = −
\
12. A cylindrical region of radius R contains a uniform magnetic field, parallel to its axis,
with magnitude that is changing linearly with time. If r is the radial distance from the
cylinder axis, the magnitude of the induced electric field outside the cylinder is
proportional to:
µ0i
r
∫ E ⋅ ds = µ i ⇒ E = 2πr
0