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1. A particle having a charge q experiences a force ๐ ๐ = q (- ๐ฃ + ๐ค)N in a magnetic field ๐ when it has a velocity ๏ฎ๐ = 1 ๐ข m/s. the force becomes ๐ = q (๐ข - ๐ค) N when the velocity is changed to ๏ฎ๐ = 1 ๐ฃ m/s. The magnetic field at that point is (a) (๐ข + ๐ฃ - ๐ค) T (b) (-๐ข - ๐ฃ + ๐ค) T (c) (๐ข - ๐ฃ - ๐ค) T (d) (๐ข + ๐ฃ + ๐ค) T 2. A particle of charge per unit mass ๏ก is released from origin with a velocity ๏ฎ = ๏ฎ0๐ข in a uniform magnetic field ๐ = - B0๐ค. If the particle passes through (0, y, 0). Then y is (a) โ ๐๏ฎ ๐ (b) ๐๐ ๏ก ๏ฎ๐ (c) ๐๐ ๏ก ๐๏ฎ๐ (d) ๐๐ ๏ก โ ๏ฎ๐ ๐๐ ๏ก 3. Two wires AO and OC carry equal currents i as shown. ๏AOC = ๏ฑ. The magnitude of magnetic field at the point P on the bisector of angle ๏ฑ at a distance r from point O is (Assume one end of both wire extends to infinity) ๏ญ๐ ๐ข (a) ๏ฑ cot ๐๏ฐ๐ซ (b) ๐ ๐๐จ๐ฌ๏ฑ ) ๐ ๏ญ๐ ๐ข (๐+ (c) (d) ๐๏ฐ๐ซ ๐ฌ๐ข๐ง ๏ฑ/๐ ๏ญ๐ ๐ข ๐๏ฐ๐ซ ๏ญ๐ ๐ข ๐๏ฐ๐ซ cot sin ๏ฑ ๐ ๏ฑ ๐ 4. A long straight wire, carrying a current i is bent at its mid point to form an angle of 45 0. At a point P, distance R from the point of bending the magnetic field is (a) ( ๐โ ๐)๏ญ๐ ๐ข (b) ๐๏ฐ๐ ( ๐+๐)๏ญ๐ ๐ข (c) ๐๏ฐ๐ ( ๐+ ๐)๏ญ ๐ ๐ข (d) ๐ ๐๏ฐ๐ ( ๐โ๐)๏ญ๐ ๐ข ๐ ๐๏ฐ๐ 5. Figure shows a cross-section of a large metal sheet carrying current along its surface. The current in a strip of width dl is kdl when k is a constant. The magnetic field at a point P at a distance x from the metal sheet is (a) ๏ญ0k (b) ๏ญ0k/2 (c) ๏ญ0k.x/2 (d) ๏ญ0kx 6. A conducting circular loop of radius r carries a current i. It is placed in a uniform magnetic field B 0 such that B0 is perpendicular to the plane of the loop. The magnetic force acting on the loop is: (a) ๏ฐB0 (b) 2๏ฐirB0 (c) zero (d) ๏ฐirB0 7. A particle of mass m and charge q moves with constant velocity ๏ฎ along +ve X-direction. It enters a region containing a uniform magnetic field B directed along the negative Z-direction, extending from x = a to x = b. The minimum value of ๏ฎ required, so that the particle can just enter the region x > b is: (a) ๐ช๐๐ ๐ช ๐โ๐ ๐ (b) ๐ฆ (c) ๐ฆ ๐ช๐๐ (d) ๐ฆ ๐ช ๐+๐ ๐ ๐๐ฆ 8. A rod CD o length b carrying a current i1 is placed in the field of a long wire AB carrying a current i2 as shown: (a) the force on the rod CD is (b) the force on the CD is ๏ญ๐ ๐๏ฐ ๏ญ๐ ๐ i i ln ( ) ๐๏ฐ 1 2 ๐ ๐ i1i2ln (1 + ) if free to move the rod CD will go ๐ (c) rotational motion only. (d) translation as well as rotational motion. 9. A particle of mass m and charge q enters a uniform magnetic field ๐ต (perpendicular to paper inward) at P with velocity ๏ฎ at an angle ๏ก and leaves the field at Q at angle ๏ข as shown. Then (a) ๏ก must be equal to ๏ข. (c) length PQ = (b) length PQ = ๐๐ฆ๏ฎ ๐ฌ๐ข๐ง ๏ก ๐๐ฆ๏ฎ ๐๐จ๐ฌ ๏ก ๐ช๐ (d) particle remains in the field for time t = ๐ช๐ ๐๐ฆ๏ก ๐ช๐ Passage: A solenoid is a wire wound closely in form of helix. Generally length of solenoid is large as compared to the transverse dimension. The magnetic field inside a very tightly wound long solenoid is uniform everywhere and is zero outside it. The field at a point P on the axis of solenoid can be obtained by superposition of fields due to large number of identical coils, all having their centre on the axis of solenoid. Using Biot-Savart law, the field at P can be given as B= ๏ญ๐ ๐๏ฐ (2๏ฐn) [sin ๏ก + sin ๏ข], where n = number of turns/length, l = length of solenoid. 10. If the solenoid is of infinite length and the point P is near one end, the magnetic field is (a) ๏ญ0ni (b) ๏ญ๐ ๐ง๐ข ๐ (c) ๏ญ๐ ๐ง๐ข ๐ (d) ๐๏ญ๐ ๐ง๐ข ๐ 11. If solenoid is of finite length and the point P is on the perpendicular bisector on its axis, the magnetic field is (a) ๏ญ๐ ๐ง๐ข๐ ๐๐ + ๐๐๐ (b) ๏ญ๐ ๐ง๐ข ๐๐ + ๐๐๐ (c) ๏ญ๐ ๐ง๐ข๐ ๐๐ + ๐๐๐ (d) 12. The variation of magnetic field with distance along the axis of solenoid is (a) (b) (c) (d) ๏ญ๐ ๐ง๐ข ๐๐ + ๐๐๐ 1. (d) Let magnetic field be Bx๐ข + By ๐ฃ - Bz๐ค, - ๐ฃ + ๐ค = By๐ค - Bz๐ฃ So ๏ By = 1, Bz = 1, ๐ข - ๐ค = - Bx ๐ค + Bz ๐ข, or F1 = q (๏ฎ๐ X ๐), In second case Bx = 1, By = 1, Bz = 1, or q (-๐ฃ + ๐ค) = q [1๐ฃ X (Bx๐ข + By๐ฃ + Bz๐ค)] q (๐ข - ๐ค) = q [1 ๐ฃ X (Bx ๐ข + By ๐ฃ + Bz ๐ค)] Hence the magnetic field is (๐ข + ๐ฃ + ๐ค) T. 2. (c) Since the angle between velocity and magnetic field is 90 0. Path of the particle will be circular and from the diagram, the radius of circular path will be r= ๐ฒ = ๐ ๐ฆ๏ฎ๐ ๐ช๐๐ ๏ฎ๐ ๐ช ๐ ๐ฆ ๐ = ๏ฎ๐ = ๐๐ ๏ก , or ๐๏ฎ ๐ y= ๐๐ ๏ก 3. (c) The direction of magnetic field due to both the wires AO and OC are out of page of paper. Due to wire OC B= = ๏ญ๐ ๐ข ๐๏ฐ๐ {sin ø1 + sin ø2}, ๏ญ๐ ๐ข ๐๏ฐ๐ซ ๐ฌ๐ข๐ง ๏ฑ/๐ ๐ ๐ ๏ญ๐ ๐ข B= (1 + cos ๏ฑ/2), ๏ฑ ๏ฑ ๐ ๐๏ฐ๐ซ ๐ฌ๐ข๐ง {sin (90 - ) + sin 90} ๐ Due to both the wires, the net magnetic field is (1 + cos ๏ฑ/2) = ๐๏ฐ๐ซ ๐ฌ๐ข๐ง ๏ฑ/๐ ๏ญ๐ ๐ข = ๏ญ๐ ๐ข ๐๏ฐ๐ซ ๐ฌ๐ข๐ง ๏ฑ/๐ (1 + cos ๏ฑ/2), 4. (a) Since point P lies on the axis of straight part OA, magnetic field at point P is zero due to part OA of wire. From ๏OPN, d = R cos 450, For part OC Since both the ends O and C are on the 0 same side of normal PN, therefore, ๏ฆ1 = - 45 and ๏ฆ2 = + 900. = ๏ญ๐ ๐ข ๐๏ฐ ๐ ๐๐จ๐ฌ ๐๐๐ = ๐ ๏ญ๐ ๐ข ๐๏ฐ๐ ( [sin (- 450) = sin 900], ๐โ ๐ ๐ )= = ๏ญ๐ ๐ข ๐ ๐ [- sin 450 + 1] = ๐๏ฐ๐ ๏ญ๐ ๐ข So B = ๐๏ญ๐ ๐ข ๐๏ฐ๐ (1 - ๐๏ฐ๐ ๐ ๐ (sin ๏ฆ1 + sin ๏ฆ2) ) ( ๐โ ๐)๏ญ๐ ๐ข ๐๏ฐ๐ 5. (b) Consider two strips A and C of the sheet situated symmetrically on the two sides of P. The magnetic field at P due to strip A is B1 and that due to strip C is B2. The resultant of these two is parallel to the width AC of the sheet. The field due to the whole sheet will also be in the same direction. The field on the opposite side of the sheet at the same distance will also be of the same magnitude but in opposite direction. Applying Ampereโs law to the rectangle shown we get 2Bl = ๏ญ0 kl, B = ๏ญ0k/2 6. (c) A loop of any shape and size does not experience any force in a uniform magnetic field perpendicular to the plane of the loop. Force on an element of width d๏ฑ at an angular displacement ๏ฑ = BiR d๏ฑ in the direction perpendicular to the element ๐๐ = dF cos ๏ฑ ๐ข + dF sin ๏ฑ ๐ข or ๐๏ฐ ๐๐ข๐ ๐๐จ๐ฌ ๏ฑ ๐ F= ๐๏ฑ ๐ข + ๐๏ฐ ๐๐ข๐ ๐ฌ๐ข๐ง ๏ฑ ๐ ๐๏ฑ ๐ฃ, F=0+0=0 7. (b) Let r = radius of circular path of particle, If the particle cannot enter the region x > b, r > (b โ a), So, the enter in the region x > b ๐ฆ๐ฏ or or ๏ฎ> ๐ช๐ (๐โ๐) ๐ฆ , (b โ a) ๏ณ r ๏ฎmin = > (b โ a), ๐ช๐ ๐ช๐ (๐โ๐) ๐ฆ 8. (b) and (d) Consider an element dx of the rod at a distance x from the wire. Magnetic field due to wire AB at the element So, net force F = ๐๐ = ๏ญ ๐ ๐ข๐ ๐ข๐ ๐๏ฐ B= ๏ญ ๐ ๐ข๐ ๐๏ฐ๐ฑ , ๐+๐ ๐๐ฑ/๐ฑ, ๐ Hence choice (b) is correct. Hence, force on element dF = Bidl = F= ๏ญ ๐ ๐ข๐ ๐ข๐ ๐๏ฐ ๏ญ ๐ ๐ข๐ ๐ข๐ ๐๏ฐ๐ฑ dx ln (1 + b/a) and acts vertically up. In this problem as B is not uniform, the force is different for different elements of the rod. The force acting on the rod is maximum for the element closest to the wire and minimum for the farthest. As a result, the rod will translate as well as rotate. Choice (d) is correct. 9. (c) and (d) As the particle enters the magnetic field, it will travel in a circular path. The centre will be on line perpendicular to its velocity. The direction of q (๐ฏX๐) shows that the centre will be outside the field as shown. As ๏APO = 900, So ๏OPQ = 90 - ๏ก, Since So ๏CQB = ๏ก, Hence ๏ก = ๏ข, PQ = 2OP cos ๏OPR = 2 (OP) cos (90 - ๏ก) PQ = ๐๐ฆ๐ฏ ๐ฌ๐ข๐ง ๏ก ๐ช๐ , Choice ๏OPQ, ๏OPQ = ๏OQP = 90 - ๏ก Hence choice (c) is correct and choice (b) is wrong. Angle subtended by are PQ at its centre is 2๏ก. Hence time taken inside the field t= ๐.๐๏ก ๐๏ฐ , = ๐๏ฐ ๐ฆ๏ก ๐ช๐ ๏ฐ = ๐๐ฆ๏ก ๐ช๐ , Hence, choice (d) is correct. 10. (b) According to the question ๏ก = 0 and ๏ข = ๏ฐ/2, 11. (a) ๏ก = ๏ข, B = ๏ญ0 ni sin ๏ก with sin ๏ก = So from equation (1) B= ๏ญ0 4๏ฐ (2๏ฐnl) = ๏ญ0 ๐๐ 2 1 ๐ฟ2 + 4๐ 2 12. (a) By observing the expression of magnetic field for solenoid, one concludes the nature of graph similar to (a).