Tiêu đề 4.MOVING CHARGES AND MAGNETISM - Explanations.pdf ✅

1 (b) Circumference = length of the wire 2πr = L r = L 2π r = 1 π (∵ L = 2 m) Magnetic moment M = nIA = 1 × 1 × π [ 1 π ] 2 = 1 π Am2 2 (b) Net force on a current carrying loop in uniform magnetic field is zero. Hence the loop can’t translate. So, options (c) and (d) are wrong. From Fleming’s left hand rule we can see that if magnetic field is perpendicular to paper inwards and current in the loop is clockwise (as shown) the magnetic force Fm on each element of the loop is radially outwards, or the loops will have a tendency to expand. 3 (d) N = 100 turns, area πr 2 = 2 × 10−2m2 , i = 5A In north-south, torque = 0.3 Nm = τ1 When plane is in east-west direction, torque = 0.4 Nm = τ2 B = μ0Ni 2r M = NiA = 100 × 5 × 2 × 10−2 = 10 τ1 = MB sinθ ; τ2 = MB cos θ ∴ τ1 τ2 = tan θ ∴ tan θ = 0.3 0.4 τ1 2 = M2B 2 sin2 θ ; τ2 2 = M2B 2 cos2 θ ∴ τ1 2 + τ2 2 = (MB) 2 ∴ (0.09 + 0.16) = 102 B 2 B 2 = 0.25 100 ⇒ B = 0.5 10 = 0.05 T 4 (d) Due to decrease in crosses (×), induced current in outer loop is anticlockwise, i. e., from d to c and clockwise in inner loop i. e., from a → b 5 (a) L = n1 2πr1 = n22πr2 ⇒ n1r1 = n2r2 ⇒ r1 r2 = n2 n1 B = μ0ni 2r ⇒ B1 B2 = μ0n1i/2r1 μ0n2i/2r2 = n1 n2 . r2 r1 = ( n1 n2 ) 2 = 1 4 6 (d) Since θ = 90° Hence τ = NIAB = 1 × I × ( √3 4 l 2)B = √3 4 l 2B I 8 (c) Given, ig = i, G = R0; i = n i + i = (n + 1) i ∴ S = ig G i − ig = iR0 (n + 1)i − i = R0 n 9 (d) By Fleming’s left hand rule 10 (d) The resistance of arm PQRS is 3 times the resistance of arm PS. If resistance of arm PS = r, then resistance of arm PQRS = 3 r. Potential difference across P and S = i1 r = i2 × 3r Magnetic field induction at O due to current through arm PS is, B1 = μ0 4π i1 a [sin45° + sin 45°] acting perpendicular to the loop upwards. May field due to PQ and RS are equal and opposite. Therefore natural each other magnetic field due to QR. B2 = μ0 4π . I1 a [sin 45° + sin 45°] Perpendicular to the loop in downwards. ∴ Resultant magnetic field at centre B⃗ = B⃗ 1 + B⃗ 2 = 0 11 (a) l B  i i
F = μ0 4π . 6M1M2 r 4 = 6M1M2 r 4 ∴ [ In CGS system, μ0 4 π = 1] = 6×800×400 20×20×20×20 = 12 dyne. 12 (b) F = Bil = 2 × 1.2 × 0.5 = 1.2N 13 (c) Neon molecule is diatomic, so it’s net magnetic moment is zero 14 (d) Magnetic field at O due to PR B1 = μ0 4π 2i/3 r [sin 30° + sin 30°] = μ0 4π 2i 3r It is directed outside the paper. Magnetic field at O, due to PQR, B2 = 2 × μ0 4π (i/3) μ [sin30° + sin30°] = μ0 4π 2i 3r It is directed inside the paper. ∴ Resultant magnetic field at O, B = B1 − B2 =0 15 (d) M = NiA 16 (d) F = q(v × B⃗ ) = 0 as v and B⃗ are parallel 17 (c) Suppose length of each wire is l. Asquare = ( l 4 ) 2 = l 2 16 Acircle = πr 2 = π ( l 2π ) 2 = l 2 4π ∵ Magnetic moment M = iA ⇒ Msquare Mcircle = Asquare Acircle = l 2 /16 l 2/4π = π 4 18 (b) Direction of magnetic field at every point on axis of a current carrying coil remains same though magnitude varies. Hence magnetic induction for whole of the x-axis will remain positive Therefore, (c) and (d) are wrong Magnitude of magnetic field will vary will x according to the formula, B = μ0NIR 2 2(R2+x 2)3/2 Hence, at x = 0, B = μ0NI 2R and when x → ∞, B → 0 Slope of the graph will be dB dx = − 3μ0NIR 2 . x 2(R2 + x 2) 5/2 It means, at x = 0, slope is equal to zero or tangent to the graph at x = 0, must be parallel to x-axis. Hence (b) is correct and (a) is wrong 19 (d) Bqv = mv 2 r ⇒ r = mv Bq ... (i) Since particle was initially at rest and gained a velocity v due to a potential difference of V volt. So, KE of particle = 1 2 mv 2 = qV v = √ 2qV m ... (ii) From Eqs. (i) and (ii), we get r = m Bq √ 2qV m r = 1 B √ 2mV q ∴ Diameter of the circular path d = 2r = 2 B √ 2mV q 21 (b) r = mv Bq ie, r ∝ v or r1 r2 = v1 v2 = 1 3 22 (d) Since, the currents are flowing in the opposite directions, the magnetic field at a point equidistant from the two wires will be zero. Hence, the force acting on the charge at this instant will be zero. l/4 r
23 (d) F⃗ = i[I × B⃗ ] = 3.5 [10−2 î× (0.74 ĵ− 0.36 k̂) = (2.59k̂ − 1.25 ĵ) × 10−2 N. 24 (c) Magnetic field due to solenoid is directed along its axis. The charged particle projected along the axis of solenoid does experience any magnetic force. So, velocity of charged particle remains unchanged. 26 (c) τmax = NiAB = 1 × i × (πr 2) × B [2πr = L, ⇒ r = L 2π ] τmax = π i ( L 2π ) 2 B = L 2 iB 4π 27 (d) Cyclotron frequency is given by v = qB 2πm ∴ v = 1.6 × 10−19 × 6.28 × 10−4 2 × 3.14 × 1.7 × 10−27 = 0.94 × 104 ≈ 104 Hz 28 (a) When a charged particle having K.E. T is subjected to a transverse uniform magnetic field, it describes a circular path in the magnetic field without any change in its speed. Thus, the K.E. of the charged particle remains T at all times 29 (c) See the following figure 30 (d) The force between two parallel current carrying wires is independent of the radii of the wires. 31 (b) Sensitivity = NAB C 32 (a) Biot-Savart’s law in vector form is given as dB = μ0 4π i d1 × r r 3 33 (b) r = √ 2m1Ek1 Bq1 =√ 2m2Ek2 Bq2 Or Ek2 = m1 m2 q2 q1 Ek1 = 2m m × q q × 50 keV = 100 keV 34 (d) Since all the given forces are lying in plane, so the given loop is in equilibrium F4 ′′ = F4 cos φ = F2 F4 ′′′′ = F4 sinφ = F3 − F1 ⇒ F4 2 = F2 2 + (F3 − F1 ) 2 ⇒ F4 = √F2 2 + (F3 − F1 ) 2 35 (c) When particle enters perpendicularly in a magnetic field, it moves along a circular path with constant speed 36 (c) A man carrying suitable instruments will measure both electric and magnetic fields when the passes by a stationary electron. If can be solved by assuming the instruments to be at rest and the electron to be moving with velocity V in opposite direction to that of man, due to which there will be magnetic field as well as electric field 37 (b) l = 2π r or r = l/2π Area of circular loop, A = π r 2 Magnetic moment M = lA = i π r 2 = i π × l 2 /4π 2 or l = √4 π M/i 38 (b) Radius of path R = 1 B √ 2MV q ⇒ q m = 2V B2R2 ⇒ q m ∝ 1 R2 39 (d) B ∝ 1 r Hence, graph will be a rectangular hyperbola. 40 (d) For perpendicular magnetic field magnetic force is provided by the force so, mv 2 r = qvB Magnetic lines of forces i Plane perpendicular to conductor
ie, r = mv qB ... (i) As in uniform circular motion v = rω, so the angular frequency of circular motion will be given by ω = v r = qB m [Using Eq. (i)] and hence the time period T = 2π ω = 2πm qB ... (ii) Given, B = 3.534 × 10−5 T, q = 1.6 × 10−19 C, m = 9.1 × 10−31 kg, T =? From Eq. (ii), we get ∴ T = 2 × 3.14 × 9.1 × 10−31 3.534 × 10−5 × 1.6 × 10−19 = 1 × 10−6 s = 1μs 41 (a) rp rα = mp mα × qα qp = 1 4 × 2 1 = 1 2 . 42 (c) The Lorentz force acting on the current carrying conductor in the magnetic field is F = IBl sinθ Since, wire PQ is parallel to the direction of magnetic field, then θ = 0, ∴ FPQ = IBl sin 0° = 0 Also, wire QR is perpendicular to the direction of magnetic field, then θ = 90°. ∴ FQR = IBl sin90° = IBl 43 (a) mv 2 R = qvB. For proton Rp = mv qB = √2mpE qB and for deuteron Rd = √2mdE qB ⇒ Rd Rp = √ md mp = √2 ⇒ Rd = √2Rp 44 (b) Here, dl = dx = 1 cm = 10−2m; i = 10Am, r = 0.5 m dB⃗ = μ0 4π (d I⃗ ×r⃗ ) r 3 = μ0 4π i dl r 2 (î× ĵ) = μ0 4π i dl r 2 k̂ = 10−7 × 10 × 10−2 sin 90° (0.5) 2 k̂ = 4 × 10−8 k̂ T. 45 (c) Frequency f = Bq 2πm As proton, electron, Li +,He+ have same charge in magnitude and since magnetic field is also constant. So, f ∝ 1 m Among the given charged particles, Li+ has highest mass, therefore it will have minimum frequency. 46 (a) r1: r2 = 1: 2 and B1: B2 = 1: 3. We know that B = μ0 4π . 2πni r ⇒ i1 i2 = B1r1 B2r2 = 1 × 1 3 × 2 = 1 6 47 (d) Case-I x < R 2 |B⃗ | = 0 Case-II R 2 ≤ x < R ∫ B⃗ . dl⃗ = μ0l |B⃗ |2πx = μ0 [πx2 − π ( R 2 ) 2 ]J |B⃗ | = μ0J 2x (x 2 − R 2 4 ) Case-III x ≥ R ∫ B⃗ . dl⃗ = μ0l |B⃗ |2πx = μ0 [πR 2 − π ( R 2 ) 2 ]J |B⃗ | = μ0J 2x 3 2 R 2 |B⃗ | = 3μ0JR 2 8x So 48 (a) Magnetic field at the centre of a circular loop of radius R carrying current I is B = μ02πI 4πR = μ0I 2r Its magnetic moment is M = IA = I(πR 2 ) ∴ B M = μ0I 2R × 1 IπR2 = μ0 2πR2 = x [Given] When both the current and radius is doubled, the ratio becomes B ′ M′ = μ0 2π(2R)3 = 1 8 ( μ0 2πR3 ) = x 8 49 (a)