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8 Questions<strong>Solution</strong>Set # 12Physics Challenging ProblemsPublished in this Issue1.[C] Magnetic field due to infinite current carryingµ Jsheet is given by B = 0, where J is linear current2density.II IIIµ 0 J2µ 0 J2µ 0 J2µ 0 J2(a) (b)Fig. (a) and (b) represent the direction of magneticfield due to current carrying sheets. For x < a,µ 0Jµ 0J(2J)µ 0 (3J) µ 0 (4J)Bresul tant = − − +2 2 2 2For a < x < 2a,µ 0Jµ 0 (2J) µ 0 (3J) µ 0 (4J)tant = − − + = −2 2 2 2For 2a < x < 3a,µ 0Jµ 0 (2J) µ 0 (3J) µ 0 (4J)Bresul tant = + − − = 02 2 2 2So, the required curve isBBresul µ 0Oa 2a 3a 4a 5aXJ(C) This situation is similar to part (i)(D) In a uniform electric field, path can be only3. A → Q B → R C → P D → Q(A) At t = 1s, flux is increasing in the inwarddirection, hence induced e.m.f. will be inanticlockwise direction.(B) At t = 5s, there is no change in flux, so inducede.m.f. is zero(C) At t = 9s, flux is increasing in upward directionhence induced e.m.f. will be in clockwise direction.(D) At t = 15s, flux is decreasing in upwarddirection, so induced e.m.f. will be inanticlockwise direction.4.[A, B, D]Rate of work done by external agent isde/dt = BIL.dx/dt = BILv and thermal powerdissipated in resistor = eI = (BvL) I clearly both areequal, hence (A).If applied external force is doubled, the rod willexperience a net force and hence acceleration. As aresult velocity increase, hence (B).Since, I = e/ROn doubling R, current and hence required powerbecome half.Since, P = BILvHence (D)2. A → P,Q,S B → P,Q,R,SC → P,Q,R,S D → Q(A) Velocity of the particle may be constant, if forcesof electric and magnetic fields balance eachother. Then, path of particle will be straight line.Also, path of particle may be helical if magneticand electric fields are in same direction. But pathof particle cannot be circular. Path can becircular if only magnetic field is present, or ifsome other forces is present which can cancel theeffect of electric field.(B) Here, all the possibilities are possible dependingupon the combinations of the three fields.5.[A]→∧^jxzn 1 sini = n 2 sinr→1.5(µ 1×j) = 2( µ 2×j)∧∧∧1.5(a i + b j) × j = 2[(c i + d j) × j]∧=∧1 .5a k 2c ka 20 4= =c 1.5 3∧∧∧∧XtraEdge for IIT-JEE 18 APRIL 2010

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