Read the following passage and answer the …...component of earth’s total magnetic field with the...
Transcript of Read the following passage and answer the …...component of earth’s total magnetic field with the...
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CBSESolved Test Papers
PHYSICSClass XII
Chapter : Magnetic Effects of Current and Magnetism
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CBSE TEST PAPER-05
CLASS - XII PHYSICS (Magnetic Effects of Current and Magnetism)
1. What type of magnetic material is used in making permanent magnets? [1]
2. Which physical quantity has the unit wb/m2? Is it a scalar or a vector quantity? [1]
3. Define angle of dip. Deduce the relation connecting angle of dip and horizontal
component of earth’s total magnetic field with the horizontal direction.
[2]
4. A point change +q is moving with speed
perpendicular to the magnetic field B as shown
in the figure. What should be the magnitude and
direction of the applied electric field so that the
net force acting on the charge is zero?
[2]
5. The energy of a charged particle moving in a uniform magnetic field does not
change. Why?
[2]
6. In the figure, straight wire AB is fixed; white the
loop is free to move under the influence of the
electric currents flowing in them. In which
direction does the loop begin to move? Justify.
[2]
7. State two factors by which voltage sensitivity of a moving coil galvanometer
can be increased?
[2]
8. The current sensitivity of a moving coil galvanometer increases by 20% when
its resistance is increased by a factor of two. Calculate by what factor, the
voltage sensitivity changes?
[3]
9. (a) Show how a moving coil galvanometer can be converted into an ammeter?
(b) A galvanometer has a resistance 30 and gives a full scale deflection for a
current of 2mA. How much resistance in what way must be connected to
convert into?
(1) An ammeter of range 0.3A
(2) A voltammeter of range 0.2V.
[5]
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CBSE TEST PAPER-05
CLASS - XII PHYSICS (Magnetic Effects of Current and Magnetism)
[ANSWERS]
Ans 1: Material having high coercivity is used in making permanent magnets.
Ans 2: Magnetic field. It is a vector quantity.
Ans 3: cosBH
B
sin
sin
cos
BV
B
BV B
B BH
Ans 4: Force on the charge due to magnetic field = qVB sin
Since isB to the plane of paper and in words
F = qVB sin 90o
F = qVB (along OY)
Force on the charge due to electric field
F = qE
Net force on change is zero if qE = qVB
(along YO)
Ans 5: The force on a charged particle in a uniform magnetic field always acts in a
direction perpendicular to the motion of the charge. Since work done by the
magnetic field on the charge is zero, hence energy of the charged particle will not
change.
Ans 6: Since current in AB and arm PQ are in same direction therefore wire will attract
the arm PQ with a force (say F1)
But repels the arm RS with a force (say F2)
Sine arm PQ is closer to the wire AB
F1 > F2 i.e. the loop will move towards the wire.
BVTan
BH
E = VB
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Ans 7: Voltage sensitivity = nBA
kR
It can be increased by
(1) increasing B using powerful magnets
(2) decreasing k by using phosphor borne strip
Ans 8: Current sensitivity ( )nBA
iI k
Voltage sensitivity ( )nBA
iiV kR
Resistance of a galvanometer increases when n and A are changed
Given R = 2R
Then n = n and A = A
New current sensitivity
' ' '( )
'
n A Biii
I k
New voltage sensitivity
' ' ' '( )
' ' 2
n A Biv
V I R kR
' 120( )
' 100Since v
I I
From (i) and (iii)
' ' 120
100
' ' 120
100
n A B
R I
n A B nAB
k k
Using equation (iv)
' 6
5 2
' 3
5
nAB
V kR
nAB
V kR
Thus voltage sensitivity decreases by a factor of 3
5.
Ans 9: (a) A galvanometer can be converted into an
ammeter by connecting a low resistance
called shunt parallel to the
galvanometer.
Since G and RS are in parallel voltage
across then is same IgRG = (I – Ig) RS
n’A’=6
5nA
' 3
5V V
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(b) (1) I = 0.3A G = 30 Ig = 2mA = 210-3 A
Sheent (S) = IgG
I Ig
3
3
2 10 30
(0.3 2 10 )S
(2) G = 30 , Ig = 2mA = 210-3A, V = 0.2V
Shunt Resistance (R) V
GIg
3
0.230
2 10R
G
IgRs R
I Ig
S = 0.2
R = 70
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