1. In the above problem the intensity of magnetisation will be (in A/m)-
(A) 7.9×102 (B) 7.9×102 (C) 7.9×102 (D) 7.9×105
Solution: (D) I – XH = 158.2 × 5000 = 7.9 × 106 Amp
2. 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-
(A) 60o (B) 45o (C) 90o (D) 30o
Solution: (D) tan Φ = Bv/BH = Bv / √3 Bv
= 1/√3 = tan30o / Φ = 30o
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3. The horizontal component of earth's magnetic field at any place is 0.36×10–4 Weber/m2. If the angle of dip at that place is 60o then the value of vertical component of earth's magnetic field will be-(in Wb/m2)-
(A) 0.12 × 10–4 (B) 0.24 × 10–4 (C) 0.40 × 10–4 (D) 0.62 × 10–4
Solution: (D) BV = BH tan Φ = 0.36 × 10–4 tan 60o
= 0.36 × 10–4 √3 = 0.62 × 10–4 Wb / m2
4. The value of angle of dip at a place on earth is 45o. If the horizontal component of earth's magnetic field is 5×10–5 Telsa then the total magnetic field of earth will be-
(A) 5/√2 × 10–5 Tesla (B) 10/√2 × 10–5 Tesla
(C) 15/√2 × 10–5 Tesla (D) zero.
Solution: (A) B = √B2V + B2H = √B2H tan2 45o + B2H = BH√2
5√2 × 10–2 Wb / m2
5. The ratio of intensities of magnetic field, at distances x and 2x from the centre of a magnet of length 2cm on its axis, will be-
(A) 4 : 1 (B) 4 : 1 approx (C) 8 : 1 (D) 8 : 1 approx.
Solution: (D) B = m0 / 4π 2M / (l3 + x3) >> m0 / 4π 2M / x3
6. The length of a bar magnet is 10 cm and its pole strength is 10–3 Weber. It is placed in a magnetic field of induction 4π × 10–3 Tesla in a direction making an angle of 30o with the field direction. The value of orque acting on the magnet will be-
(A) 2π × 10–7 N-m (B) 2π × 10–5 N-m
(C) 0.5 × 102 N-m (D) 0.5N-m.
Solution: (A) t = MB sin θ = m / B sin θ
= 10–3 × 0.1 4π × 10–3 × 0.5 – 2π × 10–7 N – m
7. A magnetic needle of magnetic moment 60 amp-m2 experiences 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 declination will be-
(A) 30° (B) 45° (C) 60° (D) 90°
Solution: (A) t = MB sin θ
sin θ = t / MB = 1.2 × 10–3 / 60 × 40 × 10–6 = 1/2 θ = 30o
8. The ratio of total intensities of magnetic field at the equator and the poles will be-
(A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 1 : 4
Solution: (A) Total intensity of magnetic field remains constant
Be/Bp = 1 : 1
9. The magnetic flux density of magnetic field is 1.5 Wb/m2. A proton enters this field with a velocity of 2×107 m/s in a direction making an angle of 30o with the field. The force acting on the proton will be-
(A) 2.4 × 10–12 Newton (B) 0.24 × 10–12 Newton
(C) 24 × 10–12 Newton (D) 0.24 × 10–12 Newton
Solution: (A) F = qvB sin θ = 1.6 × 10–19 × 2 × 107 × 1.5 × 0.5
= 2.4 × 10–12
10. The intensity of magnetic field at a point X on the axis of a small magnet is equal to the field intensity at another point Y on its equatorial axis. The ratio of distances of X and Y from the centre of the magnet will be-
(A) (B)–3 (B) (B)–3 (C) 23 (D) 21/3
Solution: (D) m0 / 4π 2M / d3x = m0 / 4π M/d3y dx/dy = 21/3
11. Two magnets A and B are equal in length, breadth and mass, but their magnetic moments are different. If the time period of B in a vibration magnetometer is twice that of A, then the ratio of magnetic moments will be-
(A) 1/2 (B) 2 (C) 4 (D) 12
Solution: (C) T = 2π √I / MB M μ 1/T2
12. The period of oscillation of a freely suspended bar magnet is 4 second. If it is cut into two equal parts lengthwise then the time period of each part will be-
(A) 4 sec. (B) 2 sec. (C) 0.5 sec. (D) 0.25 sec.
Solution: (A) T = 2π √I / MB = 2π √ml2/ 12 × m/b = 4 sec
13. 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 magnets is the same but the magnetic moment of B is four times that of A. The time period of B will be-
(A) T/4 (B) T/2 (C) 2T (D) 4 T
Solution: (B) √MA/MB = TB/TA or TB = T × √MA/4MA = T/2
14. The value of current enclosed by a circular path of radius 0.30 cm is 9.42 ampere. The value of magnetic field along the path will be-
(A) 500 amp/m (B) 1000 Amp/m
(C) 5 × 104 Amp/m (D) Zero
Solution: (A) H = i / 2pr = 9.42 / 2 × 314 × 3 × 10–3 = 500 amp/m
15. A magnetic wire is bent at its midpoint at an angle of 60o. If the length of the wire is l and its magnetic moment is M then the magnetic moment of new shape of wire will be-
(A) 2 M (B) M (C) M/2 (D) M/√2
Solution: (C) M' = m × 1/2 = M/2
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(A) 3.14 amp-m2 (B) 31.4 amp-m2
(C) 314 amp-m2 (D) 0.314amp-m2
Solution: (B) M = μ R2Ni = 3.14 × 0.25 × 20 × 2 = 3.14 amp / m2
17. 1000 turns per meter are wound over a Rowland ring of ferromagnetic material. On passing a current of 2 ampere in the coil, a magnetic field of 10 Wb/m2 is produced in it. The magnetising force generated in the material will be-
(A) 1.2 × 10–3 A/m (B) 2.6 × 10–3 A/m
(C) 0.6 × 10–4 A/m (D) 2 × 10–3 A/m
Solution: (D) H = Ni / 2μr = ni = 103 × 2 = × 103 amp/m
18. A magnet makes 10 oscillation per minute at a place where the horizontal component of earth's magnetic filed (H) is 0.33 oersted. The time period of the magnet at a place where the value of H is 0.62 oersted will be-
(A) 4.38 S (B) 0.38 S (C) 2.38 S (D) 8.38 S
Solution: (A) T μ 1/√H T2 – T1 √H1 / H2 = 6 × √0.33/0.62 = 4.38 s
19. The magnetic induction inside a solenoid is 6.5×10–4T. When it is filled with iron medium then the induction becomes 1.4T. The relative permeability of iron will be-
(A) 1578 (B) 2355 (C) 1836 (D) 2154.
Solution: (D) μ1 = B/B0 = 1.4 / 6.5 × 10–4 = 2154