MPS/PHY-XII-2011/A9
PHYSICS
ASSIGNMENT-9
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1.
A wire kept along the north-south direction is allowed to fall freely. Will an emf be induced in
the wire?
No, because neither horizontal nor vertical component of earth’s magnetic field will be
intercepted by the wire.
2.
A train is moving with uniform speed from north to south. (i) Will any induced emf appear
across the ends of its axle? (ii) Will the answer be affected if the train moves from east to west?
(i)
Yes, emf will appear because the axle is intercepting the vertical component of
the earth’s magnetic field.
(ii)
No, here also the axle intercepts the vertical component of the earth’s magnetic
field, so emf is induced across the ends of the axle.
3.
As shown in the figure, a conducting rod AB moves parallel to X-axis in a uniform magnetic
field, pointing in the positive Z-direction. The end A of the rod gets positively charged. Is this
statement true? Give reasons.
Yes, the end A becomes positively charged. According to right hand thumb rule, the free
electrons experience magnetic force in the direction from A to B. Deficit of electrons
makes end A positive while excess of electrons makes end B negative.
4.
A magnet is moved in the direction indicated by an arrow between two coils AB and CD as
shown in the figure. Suggest the direction of current in each coil.
By Lenz’s Law, the ends of both the coils closer to the magnet will behave as south poles.
Hence the current induced in both the coils will flow clockwise when seen from the
magnet side.
5.
In the figure, a coil ‘B’ is connected to low voltage bulb L and placed parallel to another coil ‘A’
as shown. Explain the following observations: (i) Bulb lights (ii) bulb gets dimmer if the coil ‘B’
is moved upwards.
(i)
Bulb lights up due to the induced current set up in the coil B because of
alternating current in coil A.
(ii)
Bulb gets dimmer when the coil b is moved upwards because the flux linked with
coil B decreases and induced current also decreases.
6.
A coil A is connected to a voltmeter V and the other coil B to an a.c source D. If a large copper
sheet C, is placed between the two coils, how does the induced emf in the coil A change due to
current in the coil B?
When the copper sheet is placed between the coils, eddy currents are induced in it which
opposes the passage of magnetic flux. The rate of change of magnetic flu linked with the
coil A decreases. Hence the emf induced in coil A due to the change in current in coil B
also decreases.
7.
A rod closing the circuit moves along a U-shaped wire at a constant speed v under the action of
the force F. The circuit is in a uniform magnetic field perpendicular to its plane. Calculate F if
the rate of generation of heat is P.
The induced emf across the ends of the rod is E = Blv.
Current in the circuit I = E/R = Blv/R.
Magnetic force on the conductor, F’ = ilB, towards left.
As there is no acceleration, F’=F => ilB = F or I = F/Bl
As P = E I = Blv . F/Bl = Fv ∴ F = P/v
8.
Two circular coils, one of radius r and the other of radius R are placed coaxially with their
centers coinciding. For R>r, obtain an expression for the mutual inductance of the
arrangement.
Suppose a current ‘i’ flows through the outer circular coil of radius R. The magnetic field
at its center will be
B=
µ0i
2R
As the second co-axially placed coil has very small radius r, the field B may be
considered constant over its cross-sectional area. The flux linked with the smaller coil
will be
ϕ = BA = B × π r 2 =
µ 0i
2R
π r2
Therefore mutual inductance M =
9.
ϕ
i
=
µ0π r 2
2R
A copper rod of length L rotates with an angular speed ‘ω’ in a uniform magnetic field B. Find
the emf developed between the two ends of the rod. The field is perpendicular to the motion of
the rod.
Suppose the rod completes one revolution in time T. Then change of flux = B X Area
Swept
= B x πL2 where L is the length of the rod.
Induced emf = Change in flux/Time
Or E = (B X πL2)/T = BπL2f.
As f = ω/2π, therefore E =( BπL2)( ω/2π) = ½ (BL2 ω)
10.
(a)A toroidal solenoid with an air-core has an average radius of 15cm, area of cross-section 12
cm2 and 1200 turns. Obtain the self-inductance of the toroid. Ignore field variation across the
cross-section of the toroid. (b) A second coil of 300 turns is wound closely on the toroid above.
If the current in the primary coil is increased from zero to 2.0 A in 0.05 seconds, obtain the
induced emf in the second coil.
(a) The uniform magnetic field set up inside a solenoid is given by
B = µ0 ni =
µ0 Ni
2π R
∴ Total flux linked with the N turns is ϕ = NBA = N .
∴ Self inductance of the toroid is
µ0 N 2 A 4π ×10−7 × (1200) 2 ×12 × 10−4
∴L = =
=
i
2π R
2π × 0.15
ϕ
= 2.304 × 10−3 H = 2.3mH
(b) Here N1 = 1200, N2 = 300
dt= 0.05 s, di = 2.0 – 0 =2.0 A
µ0 Ni
µ N 2iA
.A = 0
2π R
2π R
Therefore the emf induced in the second coil is
ε =M
di µ0 N1 N 2 A di
=
dt
l
dt
4π ×10−7 × 1200 × 300 × 12 ×10−4 2.0
×
= 0.023V
2π × 0.15
0.05
11.
A square loop of side 12 cm with its sides parallel to x-axis and y-axis moves with a velocity of
8 cm/s in the positive x-direction in an environment containing a magnetic field in the positive
z-direction. The field is neither uniform in space nor constant in time. It has a gradient of
10-3 T/cm along the negative x-direction (i.e, it increases by 10-3 T/cm as one moves in the
negative x-direction), and it is decreasing in time at the rate of 10-3T/s. Determine the direction
and magnitude of the induced current in the loop if its resistance is 4.5 mΩ.
Here A = a2=(0.12m)2 = 144 X 10-4 m2
V=8cm/s= 0.08 m/s
Rate of change of magnetic field B with distance is
dB/dx = 10-3T/cm = (10-3T/10-2m) = 10-1T/m
Rate of change of magnetic field B with time is dB/dt = 10-3T/s
Induced emf due to change in field B with position x is given by
E1= dφ/dt = d(BA)/dt = A.(dB/dt)= A.(dx/dt)(dB/dx)= Av.(dB/dx)
= 144 X 10-4 X 0.08X 10-1V = 115.2 X 10-6 V
Induced emf due to change in field B with time t is
E2= dφ/dt = d(BA)/dt = A.(dB/dt)
=144 X 10-4X10-3V = 14.4 X10-6V
Therefore Total induced emf,
E1+E2 = (115.2+14.4)X10-6V =129.6X10-6V
As R = 4.5mΩ=4.5 X 10-3Ω
Therefore induced current , i = E/R = (129.6X10-6)/(4.5X10-3) A = 2.9 X 10-2 A
The two effects have been added up because both cause a decrease in flux in +ve
z-direction. The direction of induced current is such as to increase the flux through the
loop along the +ve z-direction. If for the observer the loop moves to the right, the
current will be seen to be anticlockwise.
12.
(a) Obtain an expression for the mutual inductance between a long straight wire and a square
loop of side a as shown in the figure. (b) Now assume that the straight wire carries a current of
50 A and the loop is moved to the right with a constant velocity, v = 10m/s. Calculate the
induced emf in the loop at the instant when x = 0.2m. Take a = 0.1 m and assume that the
loop has a large resistance.
(a) As shown in the figure, consider a rectangular strip of small width dr of the loop at
distance r from the wire.
Magnetic field at the location of the strip is
B=
µ 0i
2π R
This field points normally in to the plane of the loop.
Area of the strip ,A=adr.
Magnetic flux linked with the strip
dϕ = BA =
µ0i
adr
2π R
Total magnetic flux linked with the square loop,
ϕ = ∫ dϕ = ∫
r =x+a
r=x
=
µ0i
µ ia r = x + a 1
µ ia
adr = 0 ∫
dr = 0 [ln r ]xx + a
r
=
x
2π r
2π
r
2π
µ0ia
µ ia x + a µ0ia
a
[ln( x + a) − ln x] = 0 ln
=
ln(1 + )
2π
2π
x
2π
x
(b) The square loop is moving in a non-uniform magnetic field. The magnetic flux
ϕ=
Induced emf set up in the loop,
dϕ
dϕ dx
dϕ
=−
= −v
dt
dx dt
dx
=
−v
ε =−
µ0ia 1  a  µ0 a 2 v
d  µ0ia  a  
ln
1
+
=
−
v
i


−  =
dx  2π
2π  a   x 2  2π x( x + a )
 x 
1 + 
 x
Putting the values we get
13.
µ0ia
a
ln(1 + )
2π
x
linked with the loop at any instant is
ε = 1.7 X10-5V
An aircraft with a wing span of 40 m flies with a speed of 1080 km/h in the eastward direction
at a constant altitude in the northern hemisphere, where the vertical component of earth’s
magnetic field is 1.75 X 10-5 T. Find the emf that develops between the tips of the wings.
The metallic part between the wing-tips can be treated as a single conductor cutting flux-lines
due to vertical component of earth’s magnetic field. So emf is induced between the tips of its
wings.
Here l = 40 m, BV = 1.75 X 10-5 T
V= 1080 km/h = 300m/s
ε = BVlv = 0.21 V
14.
The magnetic flux through a coil perpendicular to the plane is varying according to the relation
Φ = (5t3 + 4t2 + 2t -5)Wb. Calculate the induced current through the coil at t = 2s, if the
resistance of the coil is 5Ω.
ε = -dφ/dt|t=2s = 15.6 A