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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