Tiêu đề 12-magnetic-effect-of-curr.pdf ✅

MAGNETIC EFFECT OF CURRENT 1. (C) mv = i|l ⃗ × B⃗⃗|t = Bilt ⇒ v = Bilt M = Blq M ⇒ 0 = v 2 − 2gh ⇒ B 2 l 2q 2 M2 = 2gh 2. (B) B(x) = μ0I2 2πx , dF = BI1dx = μ0I1I2dx 2πx ⇒ F = μ0I1I2 2π ∫a 2a dx x = μ0I1I2 2π ln 2 3. (A) R = mv qB , 2R < r ⇒ V < qBR M ⇒ V < qBr 2M = qr 2M μ0ni 4. (C) The plane of motion of the particle is the z − x plane. It is a case of uniform circular motion ⇒ a⃗ ⋅ v⃗ = 0 5. (C) Magnetic moment M = qL 2m = q 2m 2 5 mR 2ω ⇒ M = 1 5 qR 2ω 6. (C) B⃗⃗ = μ0I 4πa√2 [ −iˆ √2 + kˆ √2 ] 7. (B) B⃗⃗ = μ0I1 2π×10−2 kˆ + μ0I2 2π×2×10−2 jˆ 8. (D) T = 2πm qB = 2π×1 1×1 = 2πs ; R = mv qB = 1×1 1×1 = 1 m x = 0, y = 1 2 Eyq M t 2 = 1 2 1×1 1 × π 2 = π 2 2 m z = 2R = 2 m Co-ordinates will be (0, π 2 2 , 2) m 9. (A) Force on PS = F = I(√2R)B Force on PQR = F ′ = I(2R)B ⇒ F ′ = √2F 10. (A) R = mv qB > mv √2qB . Hence particle will enter in the region where magnetic field is absent. The return path will be identical to path ABC. So required time t = 2(tAB + tBC) = 2 [ θm qB + mv/√2qB cos θV ] Also sin θ = mv/√2qB mV/qB = 1 √2 from triangle BMO Hence, t = 2 [ πm 4qB + m qB] = m 2qB [π + 4] 11. (B) Speed can change only due to work done by electrostatic force. Hence qEZ = 1 2 mv 2 ⇒ 10qZ = 1 2 mv 2 ⇒ V = √ 20qz m 12. (D) B = μ0I1 2π , let the current in PQ be i2 ⇒ dF = Bi2dx = μ0i1i2 2πx dx ⇒ net anticlockwise torque on PQ is = ∫ dF ⋅ x = ∫0 l μ0i1i2 2π dx = μ0i1i2l 2π For equilibrium μ0i1i2l 2π = mg l 2 ⇒ i2 = πmg μ0i1 13. (C) R = mV qB is decreasing as v is decreasing ⇒ it enters at P ⇒ charge is negative ∵ bending shows the direction of force
14. (B) i1 = Iθ 2π ; i2 = I(2π − θ) 2π B1 = μ0i1 2r (2r − θ) 2π and B2 = μ0i2 2r ⋅ θ 2z B1 and B2 are in opposite direction, but have same magnitude = μ0Iθ(2π−θ) 8π2r ⇒ net field is zero. 15. (A) B1 = KI a , B2 = KI 4a√2 − KI/a 4 = (√2 − 1) 4 KI a B3 = KI a/√2 = √2KI a ⇒ B3 > B1 > B2 16. (BCD) τ⃗ = m⃗⃗⃗ × B⃗⃗ ⇒ U⃗⃗ = −m⃗⃗⃗ ⋅ B⃗⃗ 17. (BD) Electric field along the axis is non zero due to P.D. along axis. 18. (BCD) Force on ab will be stronger than bc. 19. (ACD) K. E = qV = 1 2 mV 2 ⇒ R = mV qB 20. (AD) Use symmetry and Ampere's law 21. (AD) dB = μ0σ2πxdx R×2x dω 2π ; B = 1 2 ∫ μ0ω q πR2 dx 22. (ABC) ArcAB = π 3 r = πmV 3qB Time ' t ' = ( I 2π ) ( π 3 ) = T 6 = πm 3qB 23. (ABC)The particle will move along an arc which is part of a circle of radius r = mv Bq From the figure we can see r = R ∴ R = mv Bq ; T = πr/2 V = πr 2v ∵ r = R = mV Bq ∴ T = πm 2Bq 24. (CD) For cylinder B = μ0ir 2πR2 ; r < a = μ0i 2πr ; r ≥ a We can consider the given cylinder as a combination of two cylinders. One of radius ' R ' carrying current I in one direction and other of radius R/2 carrying current I/3 in both directions. At point A: B = μ0(I/3) 2π(R/2) + 0 = μ0I 3πR At point B: B = μ0(4I/3) 2(πR2) ( R 2 ) + 0 = μ0I 3πR 25. (ABC) y is speed of light x and z also have same dimension 26. (ABC) F⃗ = Il ⃗ × B⃗⃗. 27. (AD) Normal force of the rail on the wire = Bil ⇒ max force of friction at t = 0 is Bi intial l. μ = 2 × 100 5 ⋅ 1 ⋅ 3 4 = 30 N But weight = 2g = 20 N ⇒ force of static friction at t = 0 is 20 N Normal force at time t is Bil = 2 × 100 5+0.5t ⋅ 1 ⇒ normal force is decreasing ⇒ friction is also decreasing max. value of force static ⇒ When max. frictional force reduce to weight of the rod, it stats moving = 200 5+0.50t × 3 4 = 20 ⇒ 30 = 20 + 2t ⇒ t = 5sec 28. (BC) Consider the solid cylinder as super position of solenoids. 29. (ABC) R = L A , if we double radius and cross- sectional radius, then resistance will be halved. 30. (ACD) Due to sheet B = μ0k 2 = μ0(2bJ) 2 = μ0JB.
The slab is symmetrical under translation in y so field is independent of y. Also slab is Symmetric under rotation by 180∘ around Z axis, so y component of field is odd function of x. consider the ampere loop shown in diagram 2Bh = μ0(2xhJ) ∴ B = μ0Jx 31. (AD) τ = Iα I0a 2B0 = [ Ma 2 12 × 2 + M ( a 2 ) 2 × 2] α I0a 2B0 = [ Ma 2 6 + Ma 2 2 ] α I0a 2B0 = 2 3 Ma 2α α = 3I0B0 2M 32. (ABCD) (A) We have F⃗ = q[E⃗⃗ + (V⃗⃗ × B⃗⃗)] F = q[aiˆ + (xiˆ + yiˆ) × biˆ + ciˆ + dkˆ] F⃗ = q[iˆ(a + yd) − jˆ(xd) + kˆ (xc − yb)] (B) for c = 0; d = 0; y = 0 F = qaiˆ So particle will move along straight line with increasing speed (C) for c = 0; d = 0; y = 0 F = q[(a + yd)i − ybkˆ] |F| = q√(a + yb) 2 + (yb) 2 So particle will moves along helix of varying pitch (D) Here a = 0, v⃗ ⋅ B⃗⃗ = 0 And v is perpendicular to B so particle will move in circular both. 33. (AB) 34. (AC) T = 2πm Bq = 2π αB0 At t = π αB0 = T 2 ; velocity of particle is −v0iˆ + v0kˆ Speed will always remains constant = v0√2 At t = 2π αB0 = T; displacement is equal to pitch, Δx = V0T = 2πV0 αB0 At t = 2π αB0 = T; distance = speed × T = 2√2V0π αB0 35. (ABC) θ = Arc Radius ; R = 6.28 2π × 3 R = 3 B = μ0Iα 4πR Solving B = 40π 9 × 10−7T 36. (A) First particle will travel along parabolic path OA. Let time from O to A is t. ay = −qE m x = √3mv 2 qE = (2vcos 60∘ )t0 ⇒ t0 = √3mv qE vy = uy + ayt0 = 2vsin 60∘ , − qE m √3mv qE = 0 Hence at point A, velocity will be purely along x- axis and it will be 2vcos 60∘ = v. 37. (B) Now magnetic field is switched on along y- axis. Now its path will be helical as shown below with increasing pitch towards negative y-axis. r = mv qB x = x0 + rsin θ = (2vcos 60∘ )t0 + mv qB sin ωt = v√3 mv qE + mv qB sin ( qB m t)
38. (C) z-coordinate : z = −(r − rcos θ) = − mv qB [1 − cos ( qB m t)] 39. (C) In triangle PMC cos 53∘ = MP MC 3 5 = R 4 − R 12 = 8R R = 3 2 m(R is the maximum radius of half − circle ) Rmax = mumax qB ⇒ umax = 3 m/s R = 3 2 m ( R is the maximum radius of half - circle) 40. (B) R = mu qB = 24m Let, ∠MPQ = θ By geometry ∠CPO = (37 − θ) In △ CPO OC sin (∠CPO) = OP sin (∠PCO) 20 sin (37∘ − θ) = 24 sin (180∘ − 37∘) ⇒ 5 sin (37∘ − θ) = 5 × 6 3 ⇒ sin (37∘ − θ) = 1 2 θ = 7π 180 rad. ⇒ ω = qB m ⇒ ω = 2rad/sec. ⇒ t = 7π 360 sec. 41. (D) Since there is no current passing through circular path, the integral ∮ B⃗⃗ ⋅ dl⃗⃗ along the dotted circle is zero. 42. (B) Let segment OB = OC and arc BC is a circular arc with centre at origin. Since the shown closed path ABCA encloses no current, the path integral of magnetic field over this path is zero. Hence ∫A B B⃗⃗ ⋅ d̅̅̅l̅ + ∫B C B⃗⃗ ⋅ d̅̅̅l̅ + ∫C A B⃗⃗ ⋅ d̅̅̅l̅ = 0 Because B⃗⃗ is perpendicular to segment AC at all points, therefore ∫C A B⃗⃗ ⋅ d̅̅̅l̅ = 0 Hence ∫A B B⃗⃗ ⋅ d̅̅̅l̅ = ∫C B B ̅ ⋅ d̅̅̅l̅ = μ0I 2π OB(θ) 2π = μ0I 2π tan−1 1 2