Download Past Year Paper Solution AY11/12 Semester 2 PH1102/PAP112

Past Year Paper Solution AY11/12 Semester 2
PH1102/PAP112 – Field and Oscillation
1.
a.
i. The charge will be distributed in the way shown in the
figure. Since there is no Electric Field on the outer surface
of the plate, by Gauss Law, there will be no charge on the
outer surface of the plate.
ii. The capacitance is given by:
∮
b. By superposition principle:
The electric field, along the y-axis (for y > 2a) is given by:
(
)
[
(
)
( )
(
)
(
]
(
)
)
The electric field, along y-axis (for 2a > y > 0) is given by:
(
)
(
[
)
(
)
(
]
)
The potential difference between point A and B therefore is given by:
∫
∫
(
)
[ (
)
]
Note: If you notice it, the hollow does not change the potential difference, i.e. the potential
difference between point A and B will be the same with or without the hollow region. Why?
2.
a. The velocity selector, act as filter to select ions with
a certain velocity: ions with higher or lower velocity
than a given one (
), will be filtered out. Its
principle lies on the balance of electric and
magnetic force acting on the ion (therefore, the
selected ion will move in straight path, while the
others will be bent either upward or downward.
The second region is the place where the filtered ions will move with a half-circular path before
hitting the photographic plate. Since the radius of the movement depends on the mass of the
ion, the ion mass will determine the point where the ions will hit the plate, which is given by:
b. The current due to the ring’s rotation is given by:
dB
From Biot-Sarvart Law, the magnetic field in z-direction along the axis is
given by:
z
(
R
)
(
)
(
)
3.
a. The definition of EMF: line integral of force per unit charge. Therefore:
√
√
∫
√
b. The differential equation of the charge on the capacitor:
(
∫
)
(
∫
(
( )
)
(
)
(
)
(
)
)
(
)
(
)
4.
a. The electric field along the x-axis by the ring, when x is small (x << R):
( )
(
)
The equation of motion of the mass is therefore given by:
̈
[
The frequency of oscillation:
√
b. Right after the switch is closed:
Long after the switch is closed:
]
5.
a. The circuit equation:
Taking the solution to be ( )
:
√
The condition for the oscillation not to happen:
is pure real. Therefore:
√
b. After a sufficient long time, all of the energy will be dissipated away, i.e. no energy left in the
circuit. Therefore, the total energy dissipated is equal to the initial energy, which is given by:
– End of Paper –
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