1. Xtra Edge February 2012 - Career Point

2. A plane spiral coil is made on a thin insulated wire and
has N = 100 turns. Radii of inside and outside turns are
a = 10 cm and b = 20 cm respectively. A magnetic field
normal to the plane of spiral exists in the space. The
magnetic field increases at a constant rate α = 0.3
tesla/second. Calculate potential difference between the
ends of the spiral.
Sol. Since, magnetic field is increasing, therefore flux
linked with the coil is also increasing. Due to
increase in flux an emf is induced in it. Potential
difference between ends of the spiral coil is equal to
magnitude of emf induced in it.
Since there are N turns in a radial width (b – a),
therefore number of turns in the spiral coil per unit
radial width is
N
n =
( b – a)
Consider a concentric circular ring of radius x and
radial thickness dx
Number of turns in this ring = n dx.
Let at some instant magnetic field strength be B,
Flux linked with the ring, φ = πx 2 B
dφ
Emf induced in this ring = (n . dx)
dt
⎛ dB ⎞
= (n . dx) ⎜πx
⎟
⎝ dt ⎠
2
∴ Total emf induced in the spiral coil,
x=
b
2
e = ∫ ( πnαx
dx)
=
3
x=
a
1 πn α (b 3 – a 3 )
= πnα x 2 dx
Substituting value of n,
1 2 2
e = πN α(a + b + ab) = 2.2 volt Ans.
3
3. A spherical shell of radius R is filled with water.
Temperature of atmosphere is (– θ) ºC. The shell is
exposed to atmosphere and all water comes down to
0ºC and then it starts freezing from outer surface
towards the centre of the shell. Assuming shell to be
highly conducting, calculate time for whole mass of
water to freeze. Thermal conductivity of ice is K and
latent heat of its fusion is L. Density of water is ρ.
Neglect expansion during freezing.
Sol. Heat flows from water to atmosphere because
atmospheric temperature is below 0ºC. Water is
filled in spherical shell, therefore heat flows in radial
direction.
Let at some instant t, thickness of ice layer be x.
Then a concentric sphere of radius (R – x) is in
liquid form as shown in figure. Heat flows from this
sphere to atmosphere through ice layer.
To calculate rate of heat flow, first we have to
calculate thermal resistance of this ice layer.
Considering a concentric spherical shell of radius r
[(R – x) < r < R] and radial thickness dr as shown in
figure.
XtraEdge for IIT-JEE 18 FEBRUARY 2012
r
(R–x)
dr
Its thermal resistance =
2
4πr
K
∴ Total thermal resistance of ice layer
R
dr
∫ 4πKr
( R – x)
2
=
R
x
4πKR(
R – x)
Temperature just inside and outside the ice layer is
0ºC and (– θ)ºC, respectively.
∴ Temperature difference = 0 – (– q) = θº C
θ
Hence, rate of heat flow, H =
⎛ x ⎞
⎜
⎟
⎝ 4πKR(
R – x)
⎠
4πKRθ( R – x)
=
x
H
rate of freezing of mass =
L
H
∴ Rate of freezing of volume =
Lρ
But it is equal to 4π(R – x) 2 dx
.
dt
∴
4πKRθ(
R – x)
Lρx
= 4p (R – x) 2 dx
.
dt
Lρx(
R – x)
or
. dx = dt
KRθ
At instant t = 0, thickness of ice layer was equal to
x = 0 and we have to calculate time t when whole
mass has frozen or when x = R
Substituting these limits,
or t =
t
Lρ
KRθ
∫dt = ∫
0
ρ
6Kθ
LR 2
R
0
x(
R – x)
dx
Ans.