Tiêu đề 5.MAGNETISM AND MATTER - Explanations.pdf ✅

1 (b) For each part m′ = m 2 2 (c) For stable equilibrium U = −MB = −(0.4)(0.16) = −0.064 J 3 (a) Given that, the horizontal component of earth’s magnetic field BH = 0.34 × 10−4 T θ = 30° We know that, for tangent galvanometer B = BH tan θ ⇒ B = 0.34 × 10−4 × tan 30° = 1.96 × 10−5 T 5 (c) Neon is diamagnetic, hence its magnetic moment is zero. 7 (a) 8 (c) Dipole is the ultimate individual unit of magnetism in any magnet. 9 (b) Frog is leveited in magnetic field produced by the current in vertical solenoid below the frog due to repulsion, so body of frog behaves as diamagnetic substance. 10 (b) τ = MH sin θ = MH sin 30° = MH 2 11 (a) A bar magnet having N-S pole, strength m and length 2I be placed in a uniform magnetic field of strength B making an angle θ with the direction of the magnetic field. Force on N-pole of the magnet=mB (along the direction of magnetic field B.) Force on S-pole of the magnet = mB (along the direction of magnetic field B.) Force on S-pole of the magnet = mB (opposite to the direction of magnetic field B). Therefore, net magnetic force on the dipole is zero. 13 (c) A bar magnet is equivalent to a current carrying solenoid due to the following facts: (i) Both rest in north-south direction when suspended freely (ii) Both have two poles: north pole and the south pole. (iii) The like poles of both repel each other while the unlike poles attract each other. 14 (d) From the characteristic of B-H curve 15 (d) The bar magnet has coercivity 4 × 103 Am−1 ,i. e., it requires a magnetic intensity H = 4 × 103 Am−1 to get demagnetised. Let i be the current carried by solenoid having n number of turns per metre length, then by definition H = ni. Here H = 4 × 103amp turn metre −1 n = N l = 60 0.12 = 500 turn metre −1 ⇒ i = H n = 4 × 103 500 = 8.0 A 16 (c) As θ is with B, therefore according to tangent law, H = B tan θ = Bsinθ cosθ or B sinθ = Hcosθ 17 (b) As we know for circulating electron magnetic moment M = 1 2 evr ... (i) And angular momentum J = mvr ...(ii) From equation (i) and (ii), M = eJ 2m 18 (b) Torque(τ) is given by τ = |M × B| Where M is magnetic dipole moment of loop given by, M = iA ∴ M = (iπr 2 )k̂ τ = |M × B| = (πr 2 iBx) Now, torque of weight will be mgr. mgr = πr 2 iBx L S N A/2 A/2 L/2 2 S N S N S N S N L/2 2 N E W S Magnetic Axis Magnetic S N Meridian
⇒ i = mg πrBx 19 (a) The weight of upper magnet should be balanced by the repulsion between the two magnets ∴ μ0 4π . m2 r 2 = 50 gwt ⇒ 10−7 × m2 (9 × 10−6) = 50 × 10−3 × 9.8 ⇒ m = −6.64 amp × m 20 (c) tanδ = V H = V √3V = 1 √3 ∴ δ = 30° = π/ 6 rad 21 (d) At point P net magnetic field Bnet = √B1 2 + B2 2 Where B1 = μ0 4π . 2M d3 and B2 = μ0 4π . M d3 ⇒ Bnet = μ0 4π . √5M d 3 22 (d) Copper is a diamagnetic material, therefore its rod aligh itself where magnetic field is weaker and perpendicular to the direction of magnetic field there. 23 (c) Pole strength of original magnet, m = M 14 Effective distance between the poles =AB M′′ = m. 2l = M 14 × 10 = M 1.4 24 (b) Because of large permeability of soft iron, magnetic lines of force prefer to pass through it. Concentration of lines in soft iron bar increases as shown in Fig. (b). 25 (c) In equilibrium, the resultant magnetic moment will be along magnetic meridian. Let N1S1 make ∠θ with resultant tanθ = M2 M1 = M √3M = 1 √3 ∴ θ = 30° 26 (a) The deflection magnetometer is most sensitive in the null method ie, when θ = 0°. 28 (c) B2 B1 = n2 2 n1 2 = 102 5 2 = 4 B2 = 4B1 = 4 × 0.3 × 10−4T = 1.2 × 10−4T Increase in field = B2 − B1 = 0.9 × 10−4 T 29 (d) Ferromagnetic substance are strongly attracted by a magnet, show all properties of a paramagnetic substance to a much higher degree. While paramagnetic substances are feebly attracted by a magnet. When ferromagnetic substance is heated, then at a definite temperature the ferromagnetic property of the substance suddenly disappears and the substance becomes paramagnetic. The temperature above which a ferromagnetic substance becomes paramagnetic is called the curie temperature (point) of the substance. 31 (a) The magnetic potential due to a magnetic dipole at distance r is V = μ0 4π M cos θ r 2 On the right bisector(ie, on axial line), θ = 0° ∴ v = μ0 4π M r 2 or V ∝ 1 r 2 32 (a) On bending a wire its pole strength remains unchanged whereas its magnetic moment changes New magnetic moment, M′′ = m(2r) = m ( 2l π ) = 2M π 33 (c) In C.G.S. Baxial = 9 = 2M x 3 ...(i) P B2 B1 d S N 1 2 N S d
Bequatorial = M ( x 2 ) 3 = 8M x 3 ... (ii) From equation (i) and (ii), Bequatorial = 36 gauss 34 (a) On axial position Ba = μ0 4π 2Mr (r 2 − l 2) 2 ; if l < < r,then Ba = μ0 4π 2M r 3 Ba = 10−7 × 2 × 10 (0.1)3 = 2 × 10−3 T [∵ z = r] 35 (b) B = μ0 4π . 2M d 3 ⇒ B = 10−7 × 2 × 1.2 (0.1) 3 = 2.4 × 10−4T 36 (a) N1 and N2 are two null points. And BH = Horizontal component of earth’s magnetic field B = Magnetic field due to bar magnets 37 (b) T = 2π√ I MBH ⇒ T1 T2 = √ (BH)2 (BH)1 ⇒ T2 = T√ (BH)1 (BH)2 = T 2 [∵ (BH)2 = 4(BH)1] 38 (a) On applying magnetic field, domains of ferromagnetic substance align themselves in the direction of magnetic field. 41 (b) Diamagnetic will be feebly repelled. Paramagnetic will be feebly attracted. Ferromagnetic will be strongly attracted 43 (b) For each part m′′ = m 2 45 (b) A dia-magnetic liquid moves from stronger parts of magnetic field to weaker parts. Therefore the meniscus of the level of solution will fall. 47 (b) F ∝ m1m2 r 2 48 (c) Susceptibility (X) = intensity of magnetisation (I) magnetic field(B) Or I = xB ∴ I = 3 × 10−4 × 4 × 10−4 Or I = 12 × 10−8 Am−1 50 (c) Diamagnetic substances are those substances in which resultant magnetic moment in an atom is zero. A paramagnetic material tends to move from a weak magnetic field to strong magnetic field. A magnetic material is in the paramagnetic phase above its Curie temperature. Typical domain size of a ferromagnetic material is 1 mm. The susceptibility of a ferromagnetic material is χ >> 1 51 (a) Water is dia-magnetic. 52 (d) From R = 2π√I/MH; length of suspension is not involved. 53 (c) Here, 2l = 20 cm ⇒ l = 10 cm, d = 40 cm. As neutral point, H = B = μ0 4π 2 Md (d2−l 2)2 3.2 × 10−5 = 10−7 × 2M(0.4) 15 × 15 × 10−4 ∴ M = 3.2 × 15 × 15 × 10−4 × 10−5 0.8 × 10−7 = 9 m = M 2l = 9 0.2 = 45 A-m 54 (a) Torque τ = iAB sin θ , i = 0.1A, θ = 90° A = 1 2 × base × height or A = 1 2 a × a√3 2 = √3a 2 4 = √3 × (0.02) 2 4 = √3 × 10−4m2 ; θ = 90° τ = 0.1 × √3 × 10−4 × 5 × 10−2 sin 90° = 5√3 × 10−7N − m 55 (a) At neutral point, μ0 4π × m d2 = V 10−7 × m (20/100) 2 = 0.4 × 10−4 m = 0.4×10−4 25×10−7 = 16 A-m 56 (a) In first case tan θ = BV BH ...(i) Second case tan θ = BV BH cos x ...(ii) N W E S N2 BH B N1 BH B N S
From equation (i) and (ii), tan θ ′ tan θ = 1 cos x 57 (c) M = ni A = 50 × 2 × 1.25 × 10−3 = 0.125 Am2 If normal to the face of the coil makes an angle θ with the magnetic induction B, then in 1st case, torque = MB cosθ = 0.04, and in second case, Torque = MBsinθ = 0.03 ∴ MB = √(0.04) 2 + (0.03) 2 = 0.05 B = 0.05 M = 0.05 0.125 = 0.4T 58 (c) As they enter the magnetic field of the earth, they are deflected away from the equator 59 (a) In SI units, we have B = μ0(H + I) 60 (d) χm = (μr − 1) ⇒ χm = (5500 − 1) = 5499 61 (d) From figure, at equilibrium tan 60° = H F ⇒ √3 = H F ⇒ F H = 1 √3 63 (a) M = mL = 4 × 10 × 10−2 = 0.4 A × m2 64 (c) t1 = 3 = 2π√ I MR , where R resultant intensity of earth’s field. t2 = 3√2 = 2π√ I MH Divide 1 √2 = √ H R = √ Rcosδ R = √cosδ cosδ = 1 2 , δ = 60°. 65 (d) W = mB cos θ = mB cos 60° = mB × 1 2 τ = mB sin θ = mB sin60° = √3 W [∵ mB = 2W] 66 (a) NS is a magnet held vertically with its north pole on the table. P is neutral point, where NP = 20 cm, figure. Clearly, 67 (b) For equilibrium of the system torques on M1 and M2 due to BH must counter balance each other, i. e. , M1 × BH = M2 × BH. If θ is the angle between M1 and BH, then angle between M2 and BH will be (90 − θ); so M1BH sinθ = M2BH sin(90 − θ) ⇒ tan θ = M2 M1 = M 3M = 1 3 ⇒ θ = tan−1 ( 1 3 ) 68 (b) T = 2π√ I MB = 2π√ wl 2/12 Pole strength × 2l × B ∴ T ∝ √Wl ∴ T2 T1 = √ w2 w1 × l2 l1 = √ w1/2 w2 × l1/2 l1 = 1 2 ⇒ T2 = T1 2 = 0.5 sec 69 (b) For H = R cosδ ∴ R = H cosδ = B0 cos45° = √2B0 70 (a) Susceptibility of a paramagnetic substance is independent of magnetising field 71 (a) Magnetic moment of bar M = 104 J/T B = 4 × 10−5T Hence work done W = M⃗ . B⃗ = 104 × 4 × 10−5 × cos 60° = 0.2 J 72 (c) T = 2π√ I1 + I2 (M1 − M2 )BH Here M1 = M2 = M, ∴ T = ∞ 73 (c) Time period of combination T = 2π√ 2I √2M. H ... (i) x BH BH cos x BV F H 60°