B' = 0 ni' B B'= 2 Q. No. 18 A long solenoid is formed by winding 20 turns/cm. The current necessary to produce a magnetic field of 20 milli tesla inside the solenoid will be approximately 1.0 A 2.0 A 4.0 A 8.0 A Correct Answer 4 Explanation B = 0 ni 3 7 10 20 10 =4 10 I 2 10 100 I= =8A 4 Q. No. 19 A long solenoid has 800 turns per metre length of solenoid. A current of 1.6 A flows through it. The magnetic induction at the end of the solenoid on its axis is 4 16 10 tesla 4 8 10 tesla 4 32 10 tesla 4 4 10 tesla Correct Answer 2 Explanation At the end of the solenoid 0nI 4 10 800 1.6 B= = 2 2 =810 4 T 7 Q. No. 20 A toroidal solenoid has 3000 turns and a mean radius of 10 cm. It has soft iron core of relative permeability 2000. What is the magnitude of magnetic field in the core when a current of 1 A is passed through the solenoid. 1.2 T 12 T 5.6 T 4.5 T Correct Answer 2 Explanation 0nI for To radial Solenoid 2 R r 0nI 2000 4 10 3000 1 B= = 2R 1 2 10 = 12 T 7 Q. No. 21 A magnetic field always exerts a force on a charged particle never exerts a force on a charged particle exerts a force, if the charged particle is moving across the magnetic field lines exerts a force, if the charged particle is moving along the magnetic field lines Correct Answer 3 Explanation F = BqV sin

Q. No. 28 A charged particle enters a uniform magnetic field with velocity vector at an angle of 45 0 with the magnetic field. The pitch of the helical path is p. The radius of the helix will be p p 2 2p p 2 Correct Answer 2 Explanation 2m Pitch P =Vcos Bq mv P = (1) Bq 2 cos mvsin r= Bq From Eqn(1) Psin P r= = tan as = 45 0 tan 45 0 = 1 2 cos 2 P r= 2 Q. No. 29 A deuteron of kinetic energy 50 kev is describing a circular orbit of radius 0.5 metre in a plane perpendicular to magnetic fieldb. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same B is 200 kev 100 kev 50 kev 25 kev Correct Answer 2 Explanation k 1 = K.E of deiform m 1 = mass of deiform 2 2 2 B q1r k 1 = 2m 1 k 2 = K.E of proton m 2 = mass of proton q 2 = charge on proton 2 2 2 B q2r k 2 = 2m 2 q 1 = q 2 m 1 = 2m 2 k1 1 = k 2 2 k 2 =2k 1 K 1 = 50 kev K 2 = 100 kev Q. No. 30 An electron and a proton travel with equal speed in the same direction at 90 0 to a uniform magnetic field as this is switched on. They experience forces which are initially

Q. No. 35 A straight horizontal wire of mass 10 mg and length 1 m carries a current of 2 ampere. What minimum magnetic field B should be applied in the region so that the magnetic force on the wire may balance its weight. 4 2.4510 T 4 4.910 T 5 4.910 T 4 9.810 T Correct Answer 3 Explanation M = 10 10 3 10 3 kg l=1m, I =2m. BI l=mg mg B= =4.910 5 Tl Il Q. No. 36 The current in wire is directed towards east and the wire is placed in magnetic field directed towards north. The force on the wire is vertically upwards vertically downwards due south due east Correct Answer 1 Explanation According to Fleming left hand Rule Q. No. 37 A current of 3 A is flowing in a linear conductor having a length of 40 cm. The conductor is placed in a magnetic field of strength 500 gauss and makes an angle of 30 0 with the direction of the field. It experiences a force of magnitude 4 310 N 2 310 N 2 3 10 N 4 310 N Correct Answer 2 Explanation F = BIl sin 2 40 1 =5 10 3 100 2 =310 2 N Q. No. 38 A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one point. If the magnetic field is switched on in the vertical direction the tension in the string will increase will decrease may increase or decrease will remain unchanged Correct Answer 3 Explanation As magnetic field is vertical upwards or downwards Tension in the string varies accordingly. Q. No. 39 A current of 10 ampere is flowing in a wire of length 1.5 metre. A force of 15 newtons acts on it when it is placed in a uniform magnetic field of 2 tesla. The angle between

respectively. The ratio of their magnetic moments will be 6 : 2 : 1 1 : 2 : 6 1 : 4 : 36 36 : 4 : 1 Correct Answer 2 Explanation M A (Area) If A 1 : A 2 : A 3 :: 1 : 2 : 6 Then, M 1 : M 2 : M 3 :: 1 : 2 : 6 Q. No. 48 Magnetic field at the centre of the circular loop of area A is B. Then the magnetic moment of the loop will be BA A 2 0 BA A 0 2BA A none of these Correct Answer 3 Explanation A= R 2 A R= I B= 0 2R 2B A I= 0 M=IA 0 2BA A M= 0 Q. No. 49 2 2 A magnetic needle has magnetic moment of 6.7 10 A.m and moment of inertia 6 7.510 kgm. It performs 10 complete oscillations in 6.7 seconds. What is the magnitude of the magnetic field. 0.01 T 0.2 T 0.5 T 0.9 T Correct Answer 1 Explanation 1 = t+ t 0 2 0 =0 1 2 20 = 6.7 2 40 = =2.8rad/sec 2 6.7 I =MB 6 I 7.5 10 2.8 B= = M 2 6.710

Correct Answer 1 Explanation 2 ML I= 12 2 L I L'= I'= 2 4 I 4 sec =T=2 MB T'=2 I' MB T' I' I 1 = = = T'=2sec T I 4I 2 Q. No. 66 The total intensity of the Earth s magnetic field at equator is 5 units. What is its value at the poles? 5 4 3 2 Correct Answer 1 Explanation Total intensity of the earth s magnetic field is constant at every point of earth. Q. No. 67 At a certain place, horizontal component of Earth s field is 3 times the vertical component. The angle of dip at this place is 0 /3 /6 none of the above Correct Answer 3 Explanation B 4 = 3 Bv Bv 1 tan = = = tan30 B4 3 o =30 = 6 o Q. No. 68 In a magnetic meridian of a certain place, horizontal component of earth s field is 0.25 G and the angle of dip is 60 0. What is the magnetic field of the earth at this location. 0.5 G 0.25 G 0.25 3G none of these Correct Answer 1 Explanation B 4 = 0.25G =60 o B B= =0.5G cos60 o Q. No. 69 The angles of dip at the poles and the equator respectively are 30 0, 60 0

90 0, 0 0 30 0, 90 0 0 0, 0 0 Correct Answer 2 Explanation At Pole B 4 =0 cos = 0 i.e., =90 At equator B = 0 sin = 0 i.e., = 0 v o o Q. No. 70 At a certain place, the horizontal component of the earth s magnetic field is B 0 and the angle of dip is 45 0. The total intensity of the field at that place will be B 0 2 B 0 2B 0 B 2 0 Correct Answer 2 Explanation =45 o B =B 4 0 B 4 =B cos45 o B B o = 2 B= 2B o Q. No. 71 At a certain place on earth, a magnetic needle is placed along the magnetic meridian at an angle of 60 0 to the horizontal. If the horizontal component of the magnetic field at 5 the place is found to be210 T. What is the magnitude of total earth s field at that place. 4 210 T 5 410 T 10 5 T 5 310 T Correct Answer 2 Explanation B =210 T =B cos60 4 B=410 5 T 5 o Q. No. 72 Agonic line is that curve at which Total intensity of each s magnetic field is same The angle of dip is same Angle of declination is same Magnetic declination is zero Correct Answer 4 Explanation Magnetic declination is zero Q. No. 73 The magnetic lines of force due to horizontal component of earth s magnetic field will be Elliptical Circular Horizontal and parallel

Curved Correct Answer 3 Explanation Horizontal and parallel Q. No. 74 The magnetic induction along the axis of an air cored solenoid is 0.03 T. On placing an iron core inside the solenoid the magnetic induction become 1.5T. The relative permeability of iron core will be 12 40 50 300 Correct Answer 3 Explanation B 1.5 = = =50 B 0.03 0 Q. No. 75 An iron rod of length 20 cm and diameter 1 cm is placed inside a solenoid on which the number of turns is 600. The relative permeability of the rod is 1000. If a current of 0.5 A is placed in the solenoid, then the magnetization of the rod will be 2.99710 2 A/m 2.99710 3 A/m 4 2.99710 A/m 2.99710 5 A/m Correct Answer 4 Explanation N 1 4 = ni = i =3000 = 1500 L 2 I = x x +1= r 4 x = 999 I=9991500 5 I=1.4910 A /m Q. No. 76 The mass of iron rod is 80 gm and its magnetic moment is 10A.m 2. If the density of iron is 8 gm/cc, then the value of intensity of magnetization will be 10 6 A/m 3000 A/m 10 5 A/m 1 A/m Correct Answer 1 Explanation M 108000 6 I= = =10 A/m V 3 8010 Q. No. 77 A solenoid has core of a material with relative permeability 400. The winding of the solenoid are insulated from the core and carry a current of 2 ampere. If the number of turns is 1000 per meter, what is magnetic flux density inside the core?

0.4 T 0.5 T 0.7 T 1.0 T Correct Answer 4 Explanation B= 4 = r 0 4 = r B 0 =400 0 ni 7 B=400 4 10 1000 2 B=1T Q. No. 78 The magnetic susceptibility of a material of a rod is 499. Permeability of vacuum is 7 4 10 H/m. Absolute permeability of the material of the rod in henry/meter is 10 4 10 3 10 2 10 Correct Answer 4 Explanation X m = 499 r = 1+X m = 500 4 r = = r 0 =2 10 0 Q. No. 79 Magnetic susceptibility is negative for Paramagnetic material only Diamagnetic material only Ferromagnetic material only Paramagnetic and ferromagnetic materials Correct Answer 2 Explanation Paramagnetic and ferromagnetic materials Q. No. 80 A magnetizing field of 210 amp/m produces a magnetic flux density of 8tesla in an iron rod. The relative permeability of the rod will be 10 2 10 0 10 3 10 4 Correct Answer 4 Explanation B 4 = = =4 10 4 3 10 4 r = =10 0 Q. No. 81 The main difference between electric lines of force and magnetic lines of force is Electric lines of force are closed curves whereas magnetic lines are open curve Electric lines of force are open curve and magnetic lines are closed curve Magnetic field lines cut each other whereas electric lines don t Electric lines of force cut each other whereas magnetic lines of force don t cut

Correct Answer 2 Explanation Electric lines of force are open curve and magnetic lines are closed curve Q. No. 82 There are 1000 turns /m in a Rowland s ring and a current of 2A flowing in windings. The value of magnetic induction produced is found to be 1.0 T. When no core is present then magnetizing field produced in the ring will be 1000 A/m 1400 A/m 2000 A/m 2400 A/m Correct Answer 3 Explanation B B H= = =ni=10002 B ni =2000 A/m Q. No. 83 In the above problem, magnetizing field in the presence of core will be 1000 A/m 2000 A/m 2400 A/m 3200 A/m Correct Answer 2 Explanation H =2000 A/m Q. No. 84 The intensity of magnetization in the presence of core will be 1000 A/m 2.310 A /m 7.94 10 A / m 4.310 A /m Correct Answer 3 Explanation I = H.X m I = H r 1 = 2000 397.7 1 5 5 = 7.93 10 A /m 7.94 10 A /m Q. No. 85 The magnetization in the absence of the core will be 2400 A/m 2.310 A /m 7.94 10 A / m Zero Correct Answer 4 Explanation In the absence of core Intensity of Magnetization = 0 Q. No. 86 The relative permeability of the material will be 397.7 448.5 533 657 Correct Answer 1 Explanation B 1 = = r = = 397.7 4 2000 0

Is permanently demagnetized Remains Ferromagnetic Correct Answer 2 Explanation Behaves like Paramagnetic material Q. No. 91 For a diamagnetic material r 1, χ m 1 r 1, χ m 1 r 1, χ m 0 r 1, χ m 0 Correct Answer 3 Explanation 1, 0 r χ m Q. No. 92 Water is Diamagnetic Paramagnetic Ferromagnetic None of these Correct Answer 1 Explanation Diamagnetic Q. No. 93 Curie s law states that Magnetic susceptibility is inversely proportional to the absolute temperature Magnetic susceptibility is inversely proportional the square root of the absolute temperature Magnetic susceptibility is directly proportional to the absolute temperature Magnetic susceptibility does not depend on temperature Correct Answer 1 Explanation 1 χ m T Q. No. 94 The hysteresis curve is studied generally for Ferromagnetic materials Paramagnetic materials Diamagnetic materials All of these Correct Answer 1 Explanation Ferromagnetic materials Q. No. 95 The B H curve (i) and (ii) shown in figure associated with