Title 04. MOVING CHARGES AND MAGNETISM.pdf ✅

04. MOVING CHARGES AND MAGNETISM NEET PREPARATION (MEDIUM PHYSICS PAPER) Date: March 12, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on 2. () : Explana on As we know that in current carrying long solenoid the magne c field is only inside the solenoid and parallel to the axis. When, an electron is projected with uniform velocity along the axis of a current carrying long solenoid it will experience a force due to the magne c field Here, , so Hence the electron will con nue to move along the axis of the solenoid. 3. () : Explana on at centroid are numerically equal & oppo‐ site to each other. (i,e. one is into the paper and the other is towards the reader) 4. () : Explana on Since Correct op on is (2). 5. () : Explana on Torque ac ng on the loop, 6. () : Explana on If is parallel to then the par cle travels along a straight line. If then path is a circular mo‐ on and if has a component along , then path will be helical. So op on (2) is correct. 7. () : Explana on As magne c field at a point on the axis of current carrying loop is given by Where, distance on axis Now, magne c field at centre Dividing eqs. (1) and (2), we get B = μ0 I N l = 4π × 10 −7 × × 2.5 = 6.28 × 10 −4T 100 (0.5) i. e. , F = qvB sin θ θ = 0 ∘ F = 0 B→ 1 + B→ 2 = 0 B1 → due to left part B2 → due right part B1 , B2 F→ = I(→l × B→) = I[(L^i) × (2^i + 3^j − 4 ^k)] = I (4L^j + 3L^k) |F→| = 5IL ( √4 2 + 3 2 = 5) τ = M→ × B→ = niA→ × B→ v B v ⊥ B v B B = (1) μ0ir 2 2(r 2+d 2) 3/2 d = Now, B1 = = (2) μ0ir 2 2(r 2+0 2) 3/2 μ0ir 2 2r 3 = ⇒ = ( ) 3/2 ⇒ = B1 = ⇒ B1 = 250μT B B1 μ0 ir 2 × 2 × (r 2) 3/2 2(r 2 + d2) 3/2 × μ0 ir 2 B B1 r 2 r 2 + d 2 54μT B1 27 5 3 54 × 125 27

16. () : Explana on In the case of metallic rod, the charge carries flow through whole of the cross sec on. Therefore, the magne c field exists both inside as well as outside. 17. () : Explana on The windings of helix carrying currents in the same direc on. They experience an a rac ve force pulling the lower end out of mercury. circuit breaks Current becomes 'Zero' Force of a rac on Helix comes back to its posi on Circuit again gets formed.This process gets re‐ peated and so helix does 'oscilatory mo on'. So correct op on is (2). 18. () : Explana on Here, resistance of galvanometer, Let be required shunt. 19. () : Explana on Let proton of mass and charge and deutron of mass and charge K.E. is same so, or 20. () : Explana on The magne c field of a solenoid is given by, For a very long current carrying solenoid, the magne c field at the ends of a very long solenoid is given by: For a long solenoid field inside is constant. So statement A is wrong and statement B is cor‐ rect. Op on (2) is correct. 21. () : Explana on Magne c field at the centre due to current through the curved por on is into the plane of the paper into the plane of the paper Magne c field at the centre due to current through the straight por on is Since point lies on the axis of the straight por‐ on Magne c field at the centre due to current through the curved por on is into the plane of the paper into the plane of the paper Magne c field at the centre due to current through the straight por on is Since point lies on the axis of the straingt pro‐ on The resultant magne c field at the is into the plane of the paper ⇒ ⇒ ⇒ ⇒ = 0 ⇒ ⇒ G = 48Ω Ig = I = 0.04I 4 100 S S = = = = 2Ω IgG (I − Ig) 0.04I × 48 (I − 0.04I) 0.04 × 48 0.96 m qp 2m qd ⇒ r = = mv Bq p Bq = [ ∵ p = mv p = √2m KE ] √2 m K ⋅ E Bq r ∝ √m q ⇒ IS = IG 95 100 5 100 = √2 : 1 rd rp B = μonI[sin α + sin β] 1 2 α = 90 ∘ , β = 0 B = μonI = 1 2 Bcentre 2 M DA B1 = × ( ) μ0I 4πR 3π 2 = 3μ0I 8R M AB B2 = 0 M AB M BC B3 = × μ0I 4√π2R π 2 = μ0I 16R M CD B4 = 0 M CD M B = B1 + B2 + B3 + B4 = + 0 + + 0 3μ0I 8R μ0I 16R
22. () : Explana on Galvanometer can be converted into ammeter by pu ng a shunt resistance in parallel. 23. () : Explana on In non-rela vis c regime Where is radius of Dees of cyclotron. In a cy‐ clotron, the me for an ion to complete one revo‐ lu on is independent of its velocity due to an al‐ terna ng electric field with some frequency . However, as ions approach the speed of light, rel‐ a vis c effects limit further accelera on. So cor‐ rect op on is (3). 24. () : Explana on As the electron beam is not deflected, then or or As the electron moves from cathode to anode, its poten al energy at the cathode appears as its ki‐ ne c energy at the anode. If is the poten al dif‐ ference between the anode and cathode, then poten al energy of the electron at cathode . Also, kine c energy of the electron at anode . According to law of conserva on of en‐ ergy, or From Eqs.(1) and (2), we have 25. () : Explana on The force per unit length ac ng between two par‐ allel wires carrying currents and and placed a distance apart is Here, As the currents are in opposite direc ons, so is repulsive 26. () : Explana on The magne c field (2) at the centre of circular cur‐ rent carrying coil of radius and current is Similarly, if current is , then Magne c field So, resultant magne c field 27. () : Explana on The solenoid can be assumed as very large Magne c field, turns cm or turns 28. () : Explana on 29. () : Explana on G = 200 Ω, ig = 20 μA S igG = (i − ig) S i. e. i = ig ( + 1) G S ∴ 20 × 10 −3 = 20 × 10 −6 ( + 1) 200 S ⇒ S = 0.20 Ω (Mv 2/R) = QvB ⇒ vmax = QBR M R fsupply T = 2πM QB Fm = Fe Bev = Ee v = (1) E B V = eV = mv 2 1 2 mv 2 = eV 1 2 v = √ (2) 2eV m √ = or = 2eV m E B e m E2 2V B2 I1 I2 d f = μ0I1I2 2πd I1 = 1A, I2 = 3, A, d = 1 m μ0 = 4π × 10 −7TmA−1 ∴ f = = 6 × 10 −7 N m− (4π×10 −7TmA−1)(1 A)(3 A) 2π(1 m) f R I B = μ0I 2R 2I = = 2B μ02I 2R = √B 2 + (2B) 2 = √5B 2 = √5B = μ0I√5 2R B = μ0ni = 4π × 10 −7 × 10 × 1000 = 4π × 10 −3 T (∵ n = 10 / n = 1000 /m) S = igG i − ig = = = 0.1Ω 10 −3 × 1000 10 − 10 −3 1 10 Br = ( ) = 0.4T μ0I 2πr B2r = B2r = Br = (0.4) = 0.2T μ0 2π I 2r 1 2 1 2