Q.1.
A bar magnet of length 10cm and having the pole strength equal to 10- weber is kept in the magnetic field having magnetic induction (B) equal to 4π × 10-3 tesla. It makes an angle of 30° with the direction of magnetic induction (B). The value of the torque acting on the magnet is
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0%
a) 2π × 10-7 newton × metre
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50%
b)2π × 10-5 newton × metre
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0%
c)0.5 newton × metre
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50%
d)0.5 × 102 newton × metre
Q.2.
The angle of dip at place i 40.6° and the intensity of the vertical component of the earth's magnetic field V=6 × 10-5 tesla. The total intensity of earth's magnetic field at this place is
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0%
a) 7 × 10-5 tesla
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50%
b) 6 × 10-5 tesla
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50%
c) 5 × 10-5 tesla
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0%
d) 9.2 × 10-5 tesla
Q.3.
Due to earth's magnetic field, charged cosmic ray particles [ CBSE 1997]
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0%
a)requires greater kinetic energy to reach equator than the pole
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50%
b) require less kinetic energy to reach the equator than the pole
-
50%
c)can never reach the pole
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0%
d)can never reach the equator
Q.4.
The needle of dip circle when placed at a geometric pole stays long
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0%
a) south north direction only
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50%
b) east west direction only
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0%
c)vertical direction
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50%
d)none of above
Q.5.
The magnetic susceptibility for diamagnetic material is
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0%
a) small and negative
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0%
b) small and positive
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100%
c)large and positive
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0%
d)large and negative
Q.6.
The correct value of dip angle at a place is 45°. If the dip circle is rotated 45° out of the meridian, then the tangent of the angle of apparent dip is
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50%
a) 1
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50%
b) 1/√2
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0%
c) √2
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0%
d) 1/2
Q.7.
At certain place, the angle of dip is 30° and the horizontal component of earth's magnetic field is 0.5 oersted. the earth's total magnetic field ( in oersted) is
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0%
a)√3
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50%
b) 1
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50%
c)1 / √3
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0%
d)1/2
Q.8.
Two small magnets each of magnetic moment 10 Am2 are placed in end on position 0.1m apart from their centres. The force acting between them is
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0%
a) 0.6 × 107 N
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50%
b) 0.006 × 107 N
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50%
c)0.6 N
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0%
d)0.06 N
Explanation
Q.9.
At a certain place, horizontal component is √3 times the vertical component. the angle of dip at this place is
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0%
a) 0
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0%
b) π/3
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50%
c)π/6
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50%
d)none of the above
Q.10.
A steel wire of length l has a magnetic moment M. It is then bent into a semicircular arc. the new magnetic moment is
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0%
a) M
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50%
b) 2M/π
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50%
c) M/l
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0%
d) M × l
Q.11.
Two identical magnetic dipoles of magnetic moment 1.0 Am2 each, placed at sepration of 2m with their axis perpendicular to each other. The resultant magnetic field at a point midway between the dipole is
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50%
a)5 × 10-7 T
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50%
b) √5 × 10-7 T
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0%
c)10-7 T
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0%
d)2 × 10-7 T
Explanation
Q.12.
The force between two short bar magnets with magnetic moments M1 and M2 whose centres are r metre apart is 8N, when their axes are in same line. If the sepration is increased to 2r, the force between them is reduced to
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0%
a) 4N
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0%
b) 2N
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100%
c)1N
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0%
d)0.5N
Q.13.
The force experienced by a pole of strength 100 A-m at a distance of 0.2m from a short magnet of length 5cm and pole strength of 200A-m on its axial line will be
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0%
a) 2.5 × 10-2 N
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0%
b) 2.5 × 10-3 N
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100%
c)5.0 × 10-2 N
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0%
d)5.0 × 10-3 N
Explanation
Q.14.
A magnet 10 cm long has a pole strength of 12 A-m. Find the magnitude of magnetic field strength B at a point on its axis at a distance of 20 cm from it. What would be the value of B, if the point were to lie at the same distance on equatorial of magnet
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0%
a) 3.4 × 10-5 T
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100%
b) 1.4 × 10-5 T
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0%
c) 1.7 × 10-5 T
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0%
d) 0.85 × 10-5 T
Explanation
Q.15.
A magnet of moment M is lying in a magnetic field of induction B. W1 is the work done in turning it from 0° to 60 ° and W2 is the work done in turning it from 30° to 90 ° . Then
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50%
a)W2=W1
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0%
b) W2=W1/2
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50%
c) W2=2W1
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0%
d) W2=√3 W1
Q.16.
A bar magnet of magnetic moment 4.0 A-m2 is free to rotate about a vertices axis through its centre. The magnet is released from rest from east-west position. Kinetic energy of the magnet in north-south position will be [ H=25µT)
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0%
a) 10-2 J
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50%
b) 10-4 J
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50%
c)10-6 J
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0%
d)0
Q.17.
The length of bar magnet is 10cm and its pole strength is 10-3 Weber. It is placed in a magnetic field 4 π×10-3 T . in the direction making an angle 30° with electric field direction. The value of torque acting on the magnet will be
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0%
a) 2π × 10-7 N-m
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50%
b) 2π × 10-5 N-m
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0%
c)0.5 × 102 N-m
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50%
d)none of these
Q.18.
A magnet having a magnetic moment of 1.0 × 104 J/T is free to rotate in horizontal plane where a magnetic field 4 × 10-5 T exists. Find the work done in rotating the magnet slowly done in rotating the magnet slowly from a direction parallel to the field to a direction 60° from the field
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0%
a) 0.4 J
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50%
b) 2 J
-
50%
c) 0.2 J
-
0%
d) 1 J
Q.19.
A magnetic needle of magnetic moment 60 amp-m2 experience a torque of 1.2 ×10-3 N-m directed in geographical north. If the horizontal intensity of earth's magnetic field at that place is 40 Wb/m2, then the angle of direction will be
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0%
a)30°
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50%
b) 45°
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50%
c)60°
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0%
d)90°
Q.20.
A bar magnet with its poles 25 cm apart and pole strength 24.0 A-m rests with its centre on a frictionless pivot. A force F is applied on the magnet at a distance of 12cm from pivot so that it is held in equilibrium at an angle of 30° with respect to a magnetic field of induction 0.25T. The value of force F is
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0%
a) 5.62 N
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0%
b) 2.56 N
-
100%
c)6.52 N
-
0%
d)6.25 N
Q.21.
A current of 1 mp. is flowing in a coil of 10 turns and with radius 10 cm. Its magnetic moment will be
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0%
a) 0.314 A-m2
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0%
b) 3140 A-m2
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50%
c)100 A-m2
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50%
d)µ0 A-m2
Q.22.
If the radius of circular coil is doubled and the current flowing through in it is halved then new magnetic moment will be if its initial magnetic moment is 4 units
-
0%
a) 8 units
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0%
b) 4 units
-
100%
c) 2 units
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0%
d) zero
Q.23.
A short bar magnet is placed with its north pole pointing south. the neutral point is 10 cm away from the centre of magnet. If H=0.4 gauss, calculate magnetic moment of the magnet
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0%
a)2 Am2
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50%
b) 1 Am2
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50%
c)0.1 Am2
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0%
d)0.2 Am2
Q.24.
At any place on earth, the horizontal component of earth's magnetic field is √3 times the vertical component. The angle of dip at that place will be
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0%
a) 60 °
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0%
b) 45°
-
100%
c)90°
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0%
d)30°
Q.25.
The horizontal component of earth's magnetic field at any place is 0.36×10-4 Weber/mIf the angle of dip at that place is 60° then the value of vertical component of earth's magnetic field will be ( in Wb/m2)
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0%
a) 0.12 × 10-4
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0%
b) 0.24 ×10-4
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100%
c)0.40 ×10-4
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0%
d)0.62 ×10-4
Q.26.
The radius of the coil of tangent galvanometer is 16 cm. How many turns of the wire should be used if a current of 40 mA is to be produced a deflection of 45°, given horizontal component of earth's field is 0.36 ×10-4 T.
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0%
a) 458
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0%
b) 229
-
50%
c) 200
-
50%
d) 115
Explanation
Q.27.
The tangent law applied when
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0%
a) magnet is suspended in a uniform magnetic field
-
0%
b) horizontal component of earth's field is present
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0%
c) there are two magnetic filed's
-
100%
d) there are two uniform magnetic field acting perpendicular to each other
Q.28.
The period of oscillation of a freely suspended bar magnet is 4 second. If it is cut into equal parts length wise then the time period of each part will be
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0%
a)4 sec
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50%
b) 2 sec
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0%
c)0.5 sec
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50%
d)0.25 sec
Q.29.
A thin magnetic needle oscillates in a horizontal plane with a period T. It is broken into n equal parts. the time period of each part will be
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0%
a) T
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0%
b) T/n2
-
100%
c)Tn2
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0%
d)T/n
Q.30.
The time period of a small magnet in a horizontal plane is T. Another magnet B oscillates at the same place in a similar manner. The size of two magnet is same but the magnetic moment of B is four times that of A. The time period of B will be
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0%
a) T/4
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0%
b) T/2
-
50%
c)2T
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50%
d)4T