Title 04. MOVING CHARGES AND MAGNETISM.pdf ✅

BUY COURSE FROM STORE TO GET INSTANT ANSWER AND EXPLANATION OF EACH QUESTION SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 04. MOVING CHARGES AND MAGNETISM #QID# 46553 (1.) A proton, a deutron and an α − particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If rp, rd and rα denote respectively the radii of the trajectories of these particles, then (a.) rα = rd > rp (b.) rα = rd = rp (c.) rα < rd < rp (d.) rα = rp < rd ANSWER: a #QID# 46530 (2.) Consider the following statements regarding a charged particle in a magnetic field (i) starting with zero velocity, it accelerates in a direction perpendicular to the magnetic field (ii) While deflecting in the magnetic field, its energy gradually increases (iii) Only the component of magnetic field perpendicular to the direction of motion of the charged particle is effective if deflecting it (iv) Direction of deflecting force on the moving charged particle is perpendicular to its velocity. Of these statement (a.) (ii) and (iii) are correct
BUY COURSE FROM STORE TO GET INSTANT ANSWER AND EXPLANATION OF EACH QUESTION SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 (b.) (iii) and (iv) are correct (c.) (ii) . (iii) and (iv) are correct (d.) (i). (ii) and (iii) are correct ANSWER: b EXPLANATION: (b) In magnetic field, the force on charged particle F⃗ = q(v⃗ × B⃗ ). When particle is at rest in magnetic field the force on it is zero, hence no acceleration. When charged particle is deflected by the magnetic field, its speed does not change but direction of velocity changes because the deflecting force acts perpendicular to v⃗ and B⃗ . The component of magnetic field perpendicular to the direction of motion is effective in deflecting the particle. #QID# 46507 (3.) The earth’s magnetic induction at certain point is 7 × 10−5 Wb/m2 . This is to be annulled by the magnetic induction at the centre of a circular conducting loop of radius 5 cm. The required current in the loop is (a.) 0.56 A (b.) 5.6 A (c.) 0.28 A (d.) 2.8 A ANSWER: b EXPLANATION: (b)
BUY COURSE FROM STORE TO GET INSTANT ANSWER AND EXPLANATION OF EACH QUESTION SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 μ0 4π × 2πi r = H ⇒ (10−7 ) × 2 × 3.142 × i 0.05 = 7 × 10−5 ∴ i = 7 × 0.05 × 10−5 2 × 3.142 × 10−7 = 35 2 × 3.142 = 5.6 amp #QID# 46361 (4.) A electron moving with kinetic energy 6 × 10−16 J enters a field of magnetic induction 6 × 10−3 Wbm−2 at right angle to its motion. The radius of its path is (a.) 3.42 cm (b.) 4.23 cm (c.) 6.17 cm (d.) 7.7 cm ANSWER: a EXPLANATION: (a) r = mv Bq = √2Ekm Bq = √2 × 6 × 10−16 × 9 × 10−31 6 × 10−3 × 1.6 × 10−19 On solving r = 3.42 cm. #QID# 46653
BUY COURSE FROM STORE TO GET INSTANT ANSWER AND EXPLANATION OF EACH QUESTION SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 (5.) A long straight wire of radius a carries a steady current i. The current is uniformly distributed across its cross-section. The ratio of the magnetic field at a 2 and 2a is (a.) 1 4 (b.) 4 (c.) 1 (d.) 1 2 ANSWER: c EXPLANATION: (c) Current density J = i πa2 From Ampere’s circuital law ∮ B. dl = μ0 . ienclosed For r < a B × 2πr = μ0 × J × πr 2