Physics-8.UNIT 15 – ASSIGNMENT SOLUTION

UNIT 15 – ASSIGNMENT SOLUTION 1. (C) Here 2. (C, D) For a loop shaped arbitrarily, we have an infinitesimal force acting on an element of loop given by ⇒ ⇒ {∴ for a closed loop = 0} 3. (C) F = qvB A = = 1011 ms–2. The direction vertically upwards (use left hand rule). 4. (D) F = i × B = i B0 iB0 |F| = iB0 5. (A, B, D) No charge in velocity implies no acceleration i.e. no net force is acting on the proton, even under the joint influence of electric and magnetic field. The thing is possible under the following situations. Situation A: E = 0, B = 0, i.e. no field exists in the region Situation B: E = 0 i.e. no electrostatic force. B ≠ 0, i.e. the charged particle enters parallel to the field, so that net force equals to zero. Situation C: E ≠ 0, i.e. the charged particle proton must experience an electrostatic force eE and hence must accelerate. Situation D: E ≠ 0, B ≠ 0 and both shown in figure Because in such a situation Fm = evB (upwards) and Fe = eE (downwards) and if both are equal in magnitude then even the proton will suffer no change in its velocity and will continue to move along the dotted line as shown in figure. In such a situation, the velocity is 6. (A, B, D) By Work–Energy Theorem Work done = Change in K.E. ⇒ ⇒ Rate of work done by at P is ⇒ ⇒ Rate of work done by at Q is zero, because at Q 7. (A, B) ⇒ |e| = a(n2 –1) For n = ±1, e = 0 For n2 – 1 = 1 we get φ = at i.e. |e| = a ⇒ n = ± for |e| = a 8. (A, B, C, D) Velocity vector is perpendicular to magnetic field. Therefore path of the particle is a circle or radius v = v0 as the speed of the particle does not change in magnetic field. Centre of the circle is C. CP = CQ ∴ or 90 – α = 90 – β ∴ α = β Further PQ = 2PR = 2r cos (90 – α). = 2r sin α ∴ ∠PCR = ∠RCQ = α ∴ ∠PCQ = 2α arc PSQ = (2π – 2α) r = ∴ 9. (A, C, D) di = Jda ⇒ di = kr2(2πr dr) ⇒ ⇒ Further, field for r > a is and field for r < a is 10. (A, C) Apply Ampere’s Circuital Law we get B(r) = 0 {inside the tube} B(2πr) = μ0i ⇒ ⇒ B ∝ 11. (B, C) Therefore path of the particle is a circle. In magnetic field speed of particle remains constant. Therefore distance moved by the particle in time is v0t or . Magnitude of velocity is always v0. Therefore, option (d) is wrong. 12. (B) If (b – a) ≥ r (r = radius of circular path of particle) The particle cannot enter the region x > b So, to enter in the region x > b r > (b –a) or or 13. (A) is the correct option. The angular momentum L of the particle is given by L = mr2ω where ω = 2πn. ∴ Frequency Further i = q × n = Magnetic moment, M = iA = × πr2; ∴ So, 14. (B) Case of positively charged particle: Two forces are acting on the positively charged particle (a) due to electric field in the positive x direction, Force due to magnetic field. ⇒ ⇒ ⇒ This force will move the positively charged particle towards Y–axis Case of negatively charged particle: Two force are acting on the negatively charged particle (A) Due to electric field in the negative X–direction (B) due to magnetic field Same direction as that of positive charge. 15. (B) When a charged particle is moving at right angle to the magnetic field than a force acts on it which behaves as a centripetal force and moves the particle in circular motion. ∴ ∴ Similarly for second particle moving with half radius as compared to first we have ⇒ ⇒ mA vA = 2mBvB ⇒ mAvA > mBvB CMP: A wire of resistance R in the form of a semicircle lies on the top of a smooth table. A uniform magnetic find B is confined to the region as shown. The ends of the semicircle are attached to spring C & D whose other ends are fixed. If r the radius of the semicircle and K is the force constant for each spring. 16. (B) Since both rings are identical and are connected in parallel, So net force constant is Kp = 2K. 17. (D) Force applied on element = id B [angel bet is π/2] ….(i) Force applied on the semicircular wire of radius r in the direction y calculated as follows: = [ from eq. (i)] = F = 2iBr 18. (C) Force = ring constant × dilacement F = Kp × x 2iBr = 2Kx 19. (B) 20. (C) Force between two straight parallel conductors. 21. (A) Where E = Electric field strength B = Magnetic field induction 22. (B) resistance is to be connected in parallel to convert it into an ammeter. 23. (D) = 18 x 10-7 N/m 24. (B) We know 25. (B) As the directions of currents are opposite to each other, force of repulsion between them is 10–5 N. 26. (A) Current produced due to the revolution of charges = qf. i = (10–6) (5) = 5 x 10–6 amp magnetic induction at the centre of the orbit. 27. (B) We know = resistance is to be connected in parallel 28. (B) r = 0.5 x 10–10 m frequency = 5 x 10–15 cps electric current i = (1.6 x 10–19) (5 x 1015) = 8 x 10–4 amp = 0.8 mA 29. (C) Resistance R = - G = Resistance is connected is series. 30. (A) = (108 Ckg–1) v = 3 x 105 ms–1, = v sin = q r