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1. The relation between μ and H for a specimen of iron is as follows-
μ = [0.4/H + 12 × 10–4] Henry / meter
The value of H which produces flux density of 1 Tesla will be-
(A) 250 A/m (B) 500 A/m (C) 750 A/m (D) 103 A/m
Solution:(B) B = μH
or B = [0.4 / H + 12 × 10–4] H
or 1 = 0.4 + 12×10–4 H
H = 500 A/m
2. The correct curve between X and 1/T for paramagnetic materials is-
Solution: (A) X ∝ 1/T
3. The SI unit of magnetic flux is-
(A) Weber (B) Maxwel (C) Tesla (D) Gauss
Solution: (A) Φ = BA = Weber m2/m2 = Weber
4. The intensity of magnetic field due to an isolated pole of strength mp at a point distant r from it will be-
(A) mp / r2 (B) mpr2 (C) r2 / mp (D) mp / r
Solution: (A) The magnetic intensity due to an isolated pole of strength mp at a distance r = mp / r2
5. A uniform magnetic field is directed from left towards right in the plane of paper. When a piece of soft iron is placed parallel to the field. The magnetic lines of force passing through it will be-
Solution: (C) The magnetic lines of force in a ferrmagnetic material are crowded together.
6. A bar magnet of magnetic moment M is cut into two equal parts. The magnetic moment of either of the parts will be-
(A) M/2 (B) 2M (C) 2M (D) l ∝ H2
Solution: (A) M = ml M' = m l/2 = M/2
7. A loop of area 0.5 m2 is placed in a magnetic field of strength 2 Tesla in direction making an angle of 60o with the field. The magnetic flux linked with the loop will be-
(A) 1/2 Weber (B) √3/2 Weber (C) 2 Weber (D) √3 Weber
Solution: (A) Φ = BA cos θ, Φ = 2 × 1/2 cos 60o = 1/2
8. The force experienced by a pole of strength 100 A-m at a distance of 0.2m from a short magnet of length 5 cm and pole strength of 200A-m on its axial line will be
(A) 2.5×10–2 N (B) 2.5×10–3N
(C) 5.0×10–2N (D) 5.0×10–3N.
Solution: (A) F = mB = μ0 / 4π 2m'l / x3 m
= 10–7 × 2 × 200 × 0.05 × 100 / 8 × 10–3
= 2.5×10–2 N
9. The magnetic susceptibility of a paramagnetic substance is 3×10–4. It is placed in a magnetising field of 4×104 amp/m. The intensity of magnetisation will be-
(A) 3 × 108 A/M (B) 12 × 108 A/M
(C) 12 A/M (D) 24 A/M
Solution: (C) I = XH = 3×10–4×4×103 = 12 A/m
10. Volt-second is the unit of-
(A) B (B) Φ (C) I (D) x
Solution: (B) e = dΦ / dt or Φ = et = volt – sec
11. The volume susceptibility of a magnetic material is 30×10–4. Its relative permeability will be-
(A) 31 × 10–4 (B) 1.003 (C) 1.0003 (D) 29 × 10–4
Solution: (B) μr = μ/μ0 = 1 + X = 1 + 30 × 10–4 = 1.003
12. A current of 2 ampere is passed in a coil of radius 0.5 m and number of turns 20. The magnetic moment of the coil is-
(A) 0.314 A–m2 (B) 3.14 A–m2
(C) 314 A–m2 (D) 31.4 A–m2
Solution: (D) M = i R2N = 2×3.14×0.25×20 = 31.4 A–m2
13. The value of magnetic susceptibility for super-conductors is-
(A) zero (B) infinity (C) +1 (D) –1
Solution: (D) For superconductor
μr = 1 + X = 0 X = –1
14. A current of 1 ampere is flowing in a coil of 10 turns and with radius 10 cm. Its magnetic moment will be-
(A) 0.314 A-m2 (B) 3140 A-m2
(C) 100 A-m2 (D) m0 A-m2
Solution: (A) M = iA = πR2 Ni
= 3.14×0.01×10×1=0.314 Am2
15. The magnetic moment of a magnet of mass 75 gm is 9×10–7 A-m2. If the density of the material of magnet is 7.5×103 kg/m3 then intensity of magnetisation will be-
(A) 0.9 A/m (B) 0.09 A/m (C) 9 A/m (D) 90 A/m
Solution: (B) l = M/V = Md / m = 9 × 10–7 × 7.5 × 103/75 × 10–3 = 0.09 A/m
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