Page 1 34.1 CHAPTER – 34 MAGNETIC FIELD 1. q = 2 ×1.6 × 10 –19 C, ? = 3 × 10 4 km/s = 3 × 10 7 m/s B = 1 T, F = qB ? = 2 × 1.6 × 10 –19 × 3 × 10 7 × 1 = 9.6 10 –12 N. towards west. ? 2. KE = 10 Kev = 1.6 × 10 –15 J, B ? = 1 × 10 –7 T (a) The electron will be deflected towards left (b) (1/2) mv 2 = KE ? V = m 2 KE ? F = qVB & accln = e m qVB Applying s = ut + (1/2) at 2 = 2 2 e V x m qVB 2 1 ? ? = V m 2 qBx e 2 = m 2 KE m 2 qBx e 2 ? = 31 15 31 2 7 19 10 1 . 9 2 10 6 . 1 10 1 . 9 1 10 1 10 6 . 1 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? By solving we get, s = 0.0148 ˜ 1.5 × 10 –2 cm 3. B = 4 × 10 –3 T ) K ˆ ( F = [4 i ˆ + 3 j ˆ × 10 –10 ] N. F X = 4 × 10 –10 N F Y = 3 × 10 –10 N Q = 1 × 10 –9 C. Considering the motion along x-axis :– F X = quV Y B ? V Y = qB F = 3 9 10 10 4 10 1 10 4 ? ? ? ? ? ? ? = 100 m/s Along y-axis F Y = qV X B ? V X = qB F = 3 9 10 10 4 10 1 10 3 ? ? ? ? ? ? ? = 75 m/s Velocity = (–75 i ˆ + 100 j ˆ ) m/s 4. B ? = (7.0 i – 3.0 j) × 10 –3 T a ? = acceleration = (---i + 7j) × 10 –6 m/s 2 Let the gap be x. Since B ? and a ? are always perpendicular a B ? ? ? = 0 ? (7x × 10 –3 × 10 –6 – 3 × 10 –3 7 × 10 –6 ) = 0 ? 7x – 21 = 0 ? x = 3 5. m = 10 g = 10 × 10 –3 kg q = 400 mc = 400 × 10 –6 C ? = 270 m/s, B = 500 ?t = 500 × 10 –6 Tesla Force on the particle = quB = 4 × 10 –6 × 270 × 500 × 10 –6 = 54 × 10 –8 (K) Acceleration on the particle = 54 × 10 –6 m/s 2 (K) Velocity along i ˆ and acceleration along k ˆ along x-axis the motion is uniform motion and along y-axis it is accelerated motion. Along – X axis 100 = 270 × t ? t = 27 10 Along – Z axis s = ut + (1/2) at 2 ? s = 2 1 × 54 × 10 –6 × 27 10 × 27 10 = 3.7 × 10 –6 ? B X s ? ? 100 a Page 2 34.1 CHAPTER – 34 MAGNETIC FIELD 1. q = 2 ×1.6 × 10 –19 C, ? = 3 × 10 4 km/s = 3 × 10 7 m/s B = 1 T, F = qB ? = 2 × 1.6 × 10 –19 × 3 × 10 7 × 1 = 9.6 10 –12 N. towards west. ? 2. KE = 10 Kev = 1.6 × 10 –15 J, B ? = 1 × 10 –7 T (a) The electron will be deflected towards left (b) (1/2) mv 2 = KE ? V = m 2 KE ? F = qVB & accln = e m qVB Applying s = ut + (1/2) at 2 = 2 2 e V x m qVB 2 1 ? ? = V m 2 qBx e 2 = m 2 KE m 2 qBx e 2 ? = 31 15 31 2 7 19 10 1 . 9 2 10 6 . 1 10 1 . 9 1 10 1 10 6 . 1 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? By solving we get, s = 0.0148 ˜ 1.5 × 10 –2 cm 3. B = 4 × 10 –3 T ) K ˆ ( F = [4 i ˆ + 3 j ˆ × 10 –10 ] N. F X = 4 × 10 –10 N F Y = 3 × 10 –10 N Q = 1 × 10 –9 C. Considering the motion along x-axis :– F X = quV Y B ? V Y = qB F = 3 9 10 10 4 10 1 10 4 ? ? ? ? ? ? ? = 100 m/s Along y-axis F Y = qV X B ? V X = qB F = 3 9 10 10 4 10 1 10 3 ? ? ? ? ? ? ? = 75 m/s Velocity = (–75 i ˆ + 100 j ˆ ) m/s 4. B ? = (7.0 i – 3.0 j) × 10 –3 T a ? = acceleration = (---i + 7j) × 10 –6 m/s 2 Let the gap be x. Since B ? and a ? are always perpendicular a B ? ? ? = 0 ? (7x × 10 –3 × 10 –6 – 3 × 10 –3 7 × 10 –6 ) = 0 ? 7x – 21 = 0 ? x = 3 5. m = 10 g = 10 × 10 –3 kg q = 400 mc = 400 × 10 –6 C ? = 270 m/s, B = 500 ?t = 500 × 10 –6 Tesla Force on the particle = quB = 4 × 10 –6 × 270 × 500 × 10 –6 = 54 × 10 –8 (K) Acceleration on the particle = 54 × 10 –6 m/s 2 (K) Velocity along i ˆ and acceleration along k ˆ along x-axis the motion is uniform motion and along y-axis it is accelerated motion. Along – X axis 100 = 270 × t ? t = 27 10 Along – Z axis s = ut + (1/2) at 2 ? s = 2 1 × 54 × 10 –6 × 27 10 × 27 10 = 3.7 × 10 –6 ? B X s ? ? 100 a Magnetic Field 34.2 6. q P = e, mp = m, F = q P × E or ma 0 = eE or, E = e ma 0 towards west The acceleration changes from a 0 to 3a 0 Hence net acceleration produced by magnetic field B ? is 2a 0 . Force due to magnetic field = B F = m × 2a 0 = e × V 0 × B ? B = 0 0 eV ma 2 downwards 7. l = 10 cm = 10 × 10 –3 m = 10 –1 m i = 10 A, B = 0.1 T, ? = 53° F = iL B Sin ? = 10 × 10 –1 ×0.1 × 0.79 = 0.0798 ˜ 0.08 direction of F is along a direction ?r to both l and B. 8. F ? = ilB = 1 × 0.20 × 0.1 = 0.02 N For F ? = il × B So, For da & cb ? l × B = l B sin 90° towards left Hence F ? 0.02 N towards left For dc & ab ? F ? = 0.02 N downward 9. F = ilB Sin ? = ilB Sin 90° = i 2RB = 2 × (8 × 10 –2 ) × 1 = 16 × 10 –2 = 0.16 N. 10. Length = l, Current = l i ˆ B ? = B 0 T ) k ˆ j ˆ i ˆ ( ? ? = T k ˆ B j ˆ B i ˆ B 0 0 0 ? ? F = ??l × B ? = ??l i ˆ × k ˆ B j ˆ B i ˆ B 0 0 0 ? ? = ??l B 0 i ˆ × i ˆ + lB 0 i ˆ × j ˆ + lB 0 i ˆ × k ˆ = ? l B 0K ˆ – ? l B 0 j ˆ or, F ? = 2 0 2 2 B l 2 ? = 2 ? l B 0 11. i = 5 A, l = 50 cm = 0.5 m B = 0.2 T, F = ilB Sin ? = ilB Sin 90° = 5 × 0.5 × 0.2 = 0.05 N ( j ˆ ) 12. l = 2 ?a Magnetic field = B ? radially outwards Current ? ‘i’ F = i l× B = i × (2 ?a × B ? ) ? = 2 ?ai B perpendicular to the plane of the figure going inside. a 0 W E 53° d c a –lA 2A b N 2R S i ? ? X l l =50 cm 0.2 T 5A x x x x x x x x x x x x x x x x x x x x x x x x P Q B ? a i Page 3 34.1 CHAPTER – 34 MAGNETIC FIELD 1. q = 2 ×1.6 × 10 –19 C, ? = 3 × 10 4 km/s = 3 × 10 7 m/s B = 1 T, F = qB ? = 2 × 1.6 × 10 –19 × 3 × 10 7 × 1 = 9.6 10 –12 N. towards west. ? 2. KE = 10 Kev = 1.6 × 10 –15 J, B ? = 1 × 10 –7 T (a) The electron will be deflected towards left (b) (1/2) mv 2 = KE ? V = m 2 KE ? F = qVB & accln = e m qVB Applying s = ut + (1/2) at 2 = 2 2 e V x m qVB 2 1 ? ? = V m 2 qBx e 2 = m 2 KE m 2 qBx e 2 ? = 31 15 31 2 7 19 10 1 . 9 2 10 6 . 1 10 1 . 9 1 10 1 10 6 . 1 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? By solving we get, s = 0.0148 ˜ 1.5 × 10 –2 cm 3. B = 4 × 10 –3 T ) K ˆ ( F = [4 i ˆ + 3 j ˆ × 10 –10 ] N. F X = 4 × 10 –10 N F Y = 3 × 10 –10 N Q = 1 × 10 –9 C. Considering the motion along x-axis :– F X = quV Y B ? V Y = qB F = 3 9 10 10 4 10 1 10 4 ? ? ? ? ? ? ? = 100 m/s Along y-axis F Y = qV X B ? V X = qB F = 3 9 10 10 4 10 1 10 3 ? ? ? ? ? ? ? = 75 m/s Velocity = (–75 i ˆ + 100 j ˆ ) m/s 4. B ? = (7.0 i – 3.0 j) × 10 –3 T a ? = acceleration = (---i + 7j) × 10 –6 m/s 2 Let the gap be x. Since B ? and a ? are always perpendicular a B ? ? ? = 0 ? (7x × 10 –3 × 10 –6 – 3 × 10 –3 7 × 10 –6 ) = 0 ? 7x – 21 = 0 ? x = 3 5. m = 10 g = 10 × 10 –3 kg q = 400 mc = 400 × 10 –6 C ? = 270 m/s, B = 500 ?t = 500 × 10 –6 Tesla Force on the particle = quB = 4 × 10 –6 × 270 × 500 × 10 –6 = 54 × 10 –8 (K) Acceleration on the particle = 54 × 10 –6 m/s 2 (K) Velocity along i ˆ and acceleration along k ˆ along x-axis the motion is uniform motion and along y-axis it is accelerated motion. Along – X axis 100 = 270 × t ? t = 27 10 Along – Z axis s = ut + (1/2) at 2 ? s = 2 1 × 54 × 10 –6 × 27 10 × 27 10 = 3.7 × 10 –6 ? B X s ? ? 100 a Magnetic Field 34.2 6. q P = e, mp = m, F = q P × E or ma 0 = eE or, E = e ma 0 towards west The acceleration changes from a 0 to 3a 0 Hence net acceleration produced by magnetic field B ? is 2a 0 . Force due to magnetic field = B F = m × 2a 0 = e × V 0 × B ? B = 0 0 eV ma 2 downwards 7. l = 10 cm = 10 × 10 –3 m = 10 –1 m i = 10 A, B = 0.1 T, ? = 53° F = iL B Sin ? = 10 × 10 –1 ×0.1 × 0.79 = 0.0798 ˜ 0.08 direction of F is along a direction ?r to both l and B. 8. F ? = ilB = 1 × 0.20 × 0.1 = 0.02 N For F ? = il × B So, For da & cb ? l × B = l B sin 90° towards left Hence F ? 0.02 N towards left For dc & ab ? F ? = 0.02 N downward 9. F = ilB Sin ? = ilB Sin 90° = i 2RB = 2 × (8 × 10 –2 ) × 1 = 16 × 10 –2 = 0.16 N. 10. Length = l, Current = l i ˆ B ? = B 0 T ) k ˆ j ˆ i ˆ ( ? ? = T k ˆ B j ˆ B i ˆ B 0 0 0 ? ? F = ??l × B ? = ??l i ˆ × k ˆ B j ˆ B i ˆ B 0 0 0 ? ? = ??l B 0 i ˆ × i ˆ + lB 0 i ˆ × j ˆ + lB 0 i ˆ × k ˆ = ? l B 0K ˆ – ? l B 0 j ˆ or, F ? = 2 0 2 2 B l 2 ? = 2 ? l B 0 11. i = 5 A, l = 50 cm = 0.5 m B = 0.2 T, F = ilB Sin ? = ilB Sin 90° = 5 × 0.5 × 0.2 = 0.05 N ( j ˆ ) 12. l = 2 ?a Magnetic field = B ? radially outwards Current ? ‘i’ F = i l× B = i × (2 ?a × B ? ) ? = 2 ?ai B perpendicular to the plane of the figure going inside. a 0 W E 53° d c a –lA 2A b N 2R S i ? ? X l l =50 cm 0.2 T 5A x x x x x x x x x x x x x x x x x x x x x x x x P Q B ? a i Magnetic Field 34.3 13. B ? = B 0 r e r e = Unit vector along radial direction F = i( B I ? ? ? ) = ilB Sin ? = 2 2 0 d a a B ) a 2 ( i ? ? = 2 2 0 2 d a B a 2 i ? ? 14. Current anticlockwise Since the horizontal Forces have no effect. Let us check the forces for current along AD & BC [Since there is no B ? ] In AD, F = 0 For BC F = iaB upward Current clockwise Similarly, F = – iaB downwards Hence change in force = change in tension = iaB – (–iaB) = 2 iaB 15. F 1 = Force on AD = ilB inwards F 2 = Force on BC = ilB inwards They cancel each other F 3 = Force on CD = ilB inwards F 4 = Force on AB = ilB inwards They also cancel each other. So the net force on the body is 0. 16. For force on a current carrying wire in an uniform magnetic field We need, l ? length of wire i ? Current B ? Magnitude of magnetic field Since F ? = ilB Now, since the length of the wire is fixed from A to B, so force is independent of the shape of the wire. 17. Force on a semicircular wire = 2iRB = 2 × 5 × 0.05 × 0.5 = 0.25 N 18. Here the displacement vector dI = ? So magnetic for i ?t B dl ? ? = i × ?B 19. Force due to the wire AB and force due to wire CD are equal and opposite to each other. Thus they cancel each other. Net force is the force due to the semicircular loop = 2iRB 20. Mass = 10 mg = 10 –5 kg Length = 1 m ? = 2 A, B = ? Now, Mg = ilB ?B = iI mg = 1 2 8 . 9 10 5 ? ? ? = 4.9 × 10 –5 T ? 21. (a) When switch S is open 2T Cos 30° = mg ? T = ? 30 Cos 2 mg = ) 2 / 3 ( 2 8 . 9 10 200 3 ? ? ? = 1.13 dl B ? a i 2 2 d a ? d ? ? B ? l ? ? C D B A x x x x x x x x x x x x x x x x x x x x x x x x B ? a B A D ? C ? B ? l ? ? l l l b B a 5 A 5 cm ? B = 0.5 T D C B X X X X X X X X X X X X A R O / 20 cm P Q T T Page 4 34.1 CHAPTER – 34 MAGNETIC FIELD 1. q = 2 ×1.6 × 10 –19 C, ? = 3 × 10 4 km/s = 3 × 10 7 m/s B = 1 T, F = qB ? = 2 × 1.6 × 10 –19 × 3 × 10 7 × 1 = 9.6 10 –12 N. towards west. ? 2. KE = 10 Kev = 1.6 × 10 –15 J, B ? = 1 × 10 –7 T (a) The electron will be deflected towards left (b) (1/2) mv 2 = KE ? V = m 2 KE ? F = qVB & accln = e m qVB Applying s = ut + (1/2) at 2 = 2 2 e V x m qVB 2 1 ? ? = V m 2 qBx e 2 = m 2 KE m 2 qBx e 2 ? = 31 15 31 2 7 19 10 1 . 9 2 10 6 . 1 10 1 . 9 1 10 1 10 6 . 1 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? By solving we get, s = 0.0148 ˜ 1.5 × 10 –2 cm 3. B = 4 × 10 –3 T ) K ˆ ( F = [4 i ˆ + 3 j ˆ × 10 –10 ] N. F X = 4 × 10 –10 N F Y = 3 × 10 –10 N Q = 1 × 10 –9 C. Considering the motion along x-axis :– F X = quV Y B ? V Y = qB F = 3 9 10 10 4 10 1 10 4 ? ? ? ? ? ? ? = 100 m/s Along y-axis F Y = qV X B ? V X = qB F = 3 9 10 10 4 10 1 10 3 ? ? ? ? ? ? ? = 75 m/s Velocity = (–75 i ˆ + 100 j ˆ ) m/s 4. B ? = (7.0 i – 3.0 j) × 10 –3 T a ? = acceleration = (---i + 7j) × 10 –6 m/s 2 Let the gap be x. Since B ? and a ? are always perpendicular a B ? ? ? = 0 ? (7x × 10 –3 × 10 –6 – 3 × 10 –3 7 × 10 –6 ) = 0 ? 7x – 21 = 0 ? x = 3 5. m = 10 g = 10 × 10 –3 kg q = 400 mc = 400 × 10 –6 C ? = 270 m/s, B = 500 ?t = 500 × 10 –6 Tesla Force on the particle = quB = 4 × 10 –6 × 270 × 500 × 10 –6 = 54 × 10 –8 (K) Acceleration on the particle = 54 × 10 –6 m/s 2 (K) Velocity along i ˆ and acceleration along k ˆ along x-axis the motion is uniform motion and along y-axis it is accelerated motion. Along – X axis 100 = 270 × t ? t = 27 10 Along – Z axis s = ut + (1/2) at 2 ? s = 2 1 × 54 × 10 –6 × 27 10 × 27 10 = 3.7 × 10 –6 ? B X s ? ? 100 a Magnetic Field 34.2 6. q P = e, mp = m, F = q P × E or ma 0 = eE or, E = e ma 0 towards west The acceleration changes from a 0 to 3a 0 Hence net acceleration produced by magnetic field B ? is 2a 0 . Force due to magnetic field = B F = m × 2a 0 = e × V 0 × B ? B = 0 0 eV ma 2 downwards 7. l = 10 cm = 10 × 10 –3 m = 10 –1 m i = 10 A, B = 0.1 T, ? = 53° F = iL B Sin ? = 10 × 10 –1 ×0.1 × 0.79 = 0.0798 ˜ 0.08 direction of F is along a direction ?r to both l and B. 8. F ? = ilB = 1 × 0.20 × 0.1 = 0.02 N For F ? = il × B So, For da & cb ? l × B = l B sin 90° towards left Hence F ? 0.02 N towards left For dc & ab ? F ? = 0.02 N downward 9. F = ilB Sin ? = ilB Sin 90° = i 2RB = 2 × (8 × 10 –2 ) × 1 = 16 × 10 –2 = 0.16 N. 10. Length = l, Current = l i ˆ B ? = B 0 T ) k ˆ j ˆ i ˆ ( ? ? = T k ˆ B j ˆ B i ˆ B 0 0 0 ? ? F = ??l × B ? = ??l i ˆ × k ˆ B j ˆ B i ˆ B 0 0 0 ? ? = ??l B 0 i ˆ × i ˆ + lB 0 i ˆ × j ˆ + lB 0 i ˆ × k ˆ = ? l B 0K ˆ – ? l B 0 j ˆ or, F ? = 2 0 2 2 B l 2 ? = 2 ? l B 0 11. i = 5 A, l = 50 cm = 0.5 m B = 0.2 T, F = ilB Sin ? = ilB Sin 90° = 5 × 0.5 × 0.2 = 0.05 N ( j ˆ ) 12. l = 2 ?a Magnetic field = B ? radially outwards Current ? ‘i’ F = i l× B = i × (2 ?a × B ? ) ? = 2 ?ai B perpendicular to the plane of the figure going inside. a 0 W E 53° d c a –lA 2A b N 2R S i ? ? X l l =50 cm 0.2 T 5A x x x x x x x x x x x x x x x x x x x x x x x x P Q B ? a i Magnetic Field 34.3 13. B ? = B 0 r e r e = Unit vector along radial direction F = i( B I ? ? ? ) = ilB Sin ? = 2 2 0 d a a B ) a 2 ( i ? ? = 2 2 0 2 d a B a 2 i ? ? 14. Current anticlockwise Since the horizontal Forces have no effect. Let us check the forces for current along AD & BC [Since there is no B ? ] In AD, F = 0 For BC F = iaB upward Current clockwise Similarly, F = – iaB downwards Hence change in force = change in tension = iaB – (–iaB) = 2 iaB 15. F 1 = Force on AD = ilB inwards F 2 = Force on BC = ilB inwards They cancel each other F 3 = Force on CD = ilB inwards F 4 = Force on AB = ilB inwards They also cancel each other. So the net force on the body is 0. 16. For force on a current carrying wire in an uniform magnetic field We need, l ? length of wire i ? Current B ? Magnitude of magnetic field Since F ? = ilB Now, since the length of the wire is fixed from A to B, so force is independent of the shape of the wire. 17. Force on a semicircular wire = 2iRB = 2 × 5 × 0.05 × 0.5 = 0.25 N 18. Here the displacement vector dI = ? So magnetic for i ?t B dl ? ? = i × ?B 19. Force due to the wire AB and force due to wire CD are equal and opposite to each other. Thus they cancel each other. Net force is the force due to the semicircular loop = 2iRB 20. Mass = 10 mg = 10 –5 kg Length = 1 m ? = 2 A, B = ? Now, Mg = ilB ?B = iI mg = 1 2 8 . 9 10 5 ? ? ? = 4.9 × 10 –5 T ? 21. (a) When switch S is open 2T Cos 30° = mg ? T = ? 30 Cos 2 mg = ) 2 / 3 ( 2 8 . 9 10 200 3 ? ? ? = 1.13 dl B ? a i 2 2 d a ? d ? ? B ? l ? ? C D B A x x x x x x x x x x x x x x x x x x x x x x x x B ? a B A D ? C ? B ? l ? ? l l l b B a 5 A 5 cm ? B = 0.5 T D C B X X X X X X X X X X X X A R O / 20 cm P Q T T Magnetic Field 34.4 (b) When the switch is closed and a current passes through the circuit = 2 A Then ? 2T Cos 30° = mg + ilB = 200 × 10 –3 9.8 + 2 × 0.2 × 0.5 = 1.96 + 0.2 = 2.16 ? 2T = 3 2 16 . 2 ? = 2.49 ? T = 2 49 . 2 = 1.245 ˜ 1.25 22. Let ‘F’ be the force applied due to magnetic field on the wire and ‘x’ be the dist covered. So, F × l = ?mg × x ? ibBl = ?mgx ? x = mg ibBl ? ? 23. ?R = F ? ? × m × g = ilB ? ? × 10 × 10 –3 × 9.8 = 20 6 × 4.9 × 10 –2 × 0.8 ? ? = 2 2 10 2 10 8 . 0 3 . 0 ? ? ? ? ? = 0.12 ? 24. Mass = m length = l Current = i Magnetic field = B = ? friction Coefficient = ? iBl = ?mg ? B = il mg ? ? 25. (a) F dl = i × dl × B towards centre. (By cross product rule) (b) Let the length of subtends an small angle of 20 at the centre. Here 2T sin ??= i × dl × B ? 2T ? = i × a × 2 ? × B [As ??? 0, Sin ? ˜ 0] ? T = i × a × B dl = a × 2 ? Force of compression on the wire = i a B ? 26. Y = Strain Stress = ? ? ? ? ? ? ? ? ? ? ? ? ? L dl r F 2 ? Y L dl = 2 r F ? ? dl = Y L r F 2 ? ? = Y a 2 r iaB 2 ? ? ? = Y r iB a 2 2 2 ? ? So, dp = Y r iB a 2 2 2 ? ? (for small cross sectional circle) dr = ? ? ? ? 2 1 Y r iB a 2 2 2 = Y r iB a 2 2 ? S b l X P X X X X X X X X X X X X X X X X X X X 6 V ? Q l i T ? ? ? ? T Page 5 34.1 CHAPTER – 34 MAGNETIC FIELD 1. q = 2 ×1.6 × 10 –19 C, ? = 3 × 10 4 km/s = 3 × 10 7 m/s B = 1 T, F = qB ? = 2 × 1.6 × 10 –19 × 3 × 10 7 × 1 = 9.6 10 –12 N. towards west. ? 2. KE = 10 Kev = 1.6 × 10 –15 J, B ? = 1 × 10 –7 T (a) The electron will be deflected towards left (b) (1/2) mv 2 = KE ? V = m 2 KE ? F = qVB & accln = e m qVB Applying s = ut + (1/2) at 2 = 2 2 e V x m qVB 2 1 ? ? = V m 2 qBx e 2 = m 2 KE m 2 qBx e 2 ? = 31 15 31 2 7 19 10 1 . 9 2 10 6 . 1 10 1 . 9 1 10 1 10 6 . 1 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? By solving we get, s = 0.0148 ˜ 1.5 × 10 –2 cm 3. B = 4 × 10 –3 T ) K ˆ ( F = [4 i ˆ + 3 j ˆ × 10 –10 ] N. F X = 4 × 10 –10 N F Y = 3 × 10 –10 N Q = 1 × 10 –9 C. Considering the motion along x-axis :– F X = quV Y B ? V Y = qB F = 3 9 10 10 4 10 1 10 4 ? ? ? ? ? ? ? = 100 m/s Along y-axis F Y = qV X B ? V X = qB F = 3 9 10 10 4 10 1 10 3 ? ? ? ? ? ? ? = 75 m/s Velocity = (–75 i ˆ + 100 j ˆ ) m/s 4. B ? = (7.0 i – 3.0 j) × 10 –3 T a ? = acceleration = (---i + 7j) × 10 –6 m/s 2 Let the gap be x. Since B ? and a ? are always perpendicular a B ? ? ? = 0 ? (7x × 10 –3 × 10 –6 – 3 × 10 –3 7 × 10 –6 ) = 0 ? 7x – 21 = 0 ? x = 3 5. m = 10 g = 10 × 10 –3 kg q = 400 mc = 400 × 10 –6 C ? = 270 m/s, B = 500 ?t = 500 × 10 –6 Tesla Force on the particle = quB = 4 × 10 –6 × 270 × 500 × 10 –6 = 54 × 10 –8 (K) Acceleration on the particle = 54 × 10 –6 m/s 2 (K) Velocity along i ˆ and acceleration along k ˆ along x-axis the motion is uniform motion and along y-axis it is accelerated motion. Along – X axis 100 = 270 × t ? t = 27 10 Along – Z axis s = ut + (1/2) at 2 ? s = 2 1 × 54 × 10 –6 × 27 10 × 27 10 = 3.7 × 10 –6 ? B X s ? ? 100 a Magnetic Field 34.2 6. q P = e, mp = m, F = q P × E or ma 0 = eE or, E = e ma 0 towards west The acceleration changes from a 0 to 3a 0 Hence net acceleration produced by magnetic field B ? is 2a 0 . Force due to magnetic field = B F = m × 2a 0 = e × V 0 × B ? B = 0 0 eV ma 2 downwards 7. l = 10 cm = 10 × 10 –3 m = 10 –1 m i = 10 A, B = 0.1 T, ? = 53° F = iL B Sin ? = 10 × 10 –1 ×0.1 × 0.79 = 0.0798 ˜ 0.08 direction of F is along a direction ?r to both l and B. 8. F ? = ilB = 1 × 0.20 × 0.1 = 0.02 N For F ? = il × B So, For da & cb ? l × B = l B sin 90° towards left Hence F ? 0.02 N towards left For dc & ab ? F ? = 0.02 N downward 9. F = ilB Sin ? = ilB Sin 90° = i 2RB = 2 × (8 × 10 –2 ) × 1 = 16 × 10 –2 = 0.16 N. 10. Length = l, Current = l i ˆ B ? = B 0 T ) k ˆ j ˆ i ˆ ( ? ? = T k ˆ B j ˆ B i ˆ B 0 0 0 ? ? F = ??l × B ? = ??l i ˆ × k ˆ B j ˆ B i ˆ B 0 0 0 ? ? = ??l B 0 i ˆ × i ˆ + lB 0 i ˆ × j ˆ + lB 0 i ˆ × k ˆ = ? l B 0K ˆ – ? l B 0 j ˆ or, F ? = 2 0 2 2 B l 2 ? = 2 ? l B 0 11. i = 5 A, l = 50 cm = 0.5 m B = 0.2 T, F = ilB Sin ? = ilB Sin 90° = 5 × 0.5 × 0.2 = 0.05 N ( j ˆ ) 12. l = 2 ?a Magnetic field = B ? radially outwards Current ? ‘i’ F = i l× B = i × (2 ?a × B ? ) ? = 2 ?ai B perpendicular to the plane of the figure going inside. a 0 W E 53° d c a –lA 2A b N 2R S i ? ? X l l =50 cm 0.2 T 5A x x x x x x x x x x x x x x x x x x x x x x x x P Q B ? a i Magnetic Field 34.3 13. B ? = B 0 r e r e = Unit vector along radial direction F = i( B I ? ? ? ) = ilB Sin ? = 2 2 0 d a a B ) a 2 ( i ? ? = 2 2 0 2 d a B a 2 i ? ? 14. Current anticlockwise Since the horizontal Forces have no effect. Let us check the forces for current along AD & BC [Since there is no B ? ] In AD, F = 0 For BC F = iaB upward Current clockwise Similarly, F = – iaB downwards Hence change in force = change in tension = iaB – (–iaB) = 2 iaB 15. F 1 = Force on AD = ilB inwards F 2 = Force on BC = ilB inwards They cancel each other F 3 = Force on CD = ilB inwards F 4 = Force on AB = ilB inwards They also cancel each other. So the net force on the body is 0. 16. For force on a current carrying wire in an uniform magnetic field We need, l ? length of wire i ? Current B ? Magnitude of magnetic field Since F ? = ilB Now, since the length of the wire is fixed from A to B, so force is independent of the shape of the wire. 17. Force on a semicircular wire = 2iRB = 2 × 5 × 0.05 × 0.5 = 0.25 N 18. Here the displacement vector dI = ? So magnetic for i ?t B dl ? ? = i × ?B 19. Force due to the wire AB and force due to wire CD are equal and opposite to each other. Thus they cancel each other. Net force is the force due to the semicircular loop = 2iRB 20. Mass = 10 mg = 10 –5 kg Length = 1 m ? = 2 A, B = ? Now, Mg = ilB ?B = iI mg = 1 2 8 . 9 10 5 ? ? ? = 4.9 × 10 –5 T ? 21. (a) When switch S is open 2T Cos 30° = mg ? T = ? 30 Cos 2 mg = ) 2 / 3 ( 2 8 . 9 10 200 3 ? ? ? = 1.13 dl B ? a i 2 2 d a ? d ? ? B ? l ? ? C D B A x x x x x x x x x x x x x x x x x x x x x x x x B ? a B A D ? C ? B ? l ? ? l l l b B a 5 A 5 cm ? B = 0.5 T D C B X X X X X X X X X X X X A R O / 20 cm P Q T T Magnetic Field 34.4 (b) When the switch is closed and a current passes through the circuit = 2 A Then ? 2T Cos 30° = mg + ilB = 200 × 10 –3 9.8 + 2 × 0.2 × 0.5 = 1.96 + 0.2 = 2.16 ? 2T = 3 2 16 . 2 ? = 2.49 ? T = 2 49 . 2 = 1.245 ˜ 1.25 22. Let ‘F’ be the force applied due to magnetic field on the wire and ‘x’ be the dist covered. So, F × l = ?mg × x ? ibBl = ?mgx ? x = mg ibBl ? ? 23. ?R = F ? ? × m × g = ilB ? ? × 10 × 10 –3 × 9.8 = 20 6 × 4.9 × 10 –2 × 0.8 ? ? = 2 2 10 2 10 8 . 0 3 . 0 ? ? ? ? ? = 0.12 ? 24. Mass = m length = l Current = i Magnetic field = B = ? friction Coefficient = ? iBl = ?mg ? B = il mg ? ? 25. (a) F dl = i × dl × B towards centre. (By cross product rule) (b) Let the length of subtends an small angle of 20 at the centre. Here 2T sin ??= i × dl × B ? 2T ? = i × a × 2 ? × B [As ??? 0, Sin ? ˜ 0] ? T = i × a × B dl = a × 2 ? Force of compression on the wire = i a B ? 26. Y = Strain Stress = ? ? ? ? ? ? ? ? ? ? ? ? ? L dl r F 2 ? Y L dl = 2 r F ? ? dl = Y L r F 2 ? ? = Y a 2 r iaB 2 ? ? ? = Y r iB a 2 2 2 ? ? So, dp = Y r iB a 2 2 2 ? ? (for small cross sectional circle) dr = ? ? ? ? 2 1 Y r iB a 2 2 2 = Y r iB a 2 2 ? S b l X P X X X X X X X X X X X X X X X X X X X 6 V ? Q l i T ? ? ? ? T Magnetic Field 34.5 27. B ? = B 0 K ˆ l x 1 ? ? ? ? ? ? ? f 1 = force on AB = iB 0 [1 + 0]l = iB 0 l f 2 = force on CD = iB 0 [1 + 0]l = iB 0 l f 3 = force on AD = iB 0 [1 + 0/1]l = iB 0 l f 4 = force on AB = iB 0 [1 + 1/1]l = 2iB 0 l Net horizontal force = F 1 – F 2 = 0 Net vertical force = F 4 – F 3 = iB 0 l 28. (a) Velocity of electron = ? Magnetic force on electron F = e ?B (b) F = qE; F = e ?B or, qE = e ?B ? eE = e ?B or, E ? = ?B (c) E = dr dV = l V ? V = lE = l ?B ? 29. (a) i = V 0 nAe ? V 0 = nae i (b) F = ilB = nA iBl = nA iB (upwards) (c) Let the electric field be E Ee = An iB ? E = Aen iB (d) dr dv = E ? dV = Edr = E×d = d Aen iB 30. q = 2.0 × 10 –8 C B ? = 0.10 T m = 2.0 × 10 –10 g = 2 × 10 –13 g ? = 2.0 × 10 3 m/ ? R = qB m ? = 1 8 3 13 10 10 2 10 2 10 2 ? ? ? ? ? ? ? ? = 0.2 m = 20 cm T = qB m 2 ? = 1 8 13 10 10 2 10 2 14 . 3 2 ? ? ? ? ? ? ? ? = 6.28 × 10 –4 s 31. r = qB mv 0.01 = e0.1 mv …(1) r = 0.1 2e V m 4 ? ? …(2) (2) ÷ (1) ? 01 . 0 r = mv 1 . 0 e 2 0.1 mVe 4 ? ? ? = 2 4 = 2 ? r = 0.02 m = 2 cm. 32. KE = 100ev = 1.6 × 10 –17 J (1/2) × 9.1 × 10 –31 × V 2 = 1.6 × 10 –17 J ? V 2 = 31 17 10 1 . 9 2 10 6 . 1 ? ? ? ? ? = 0.35 × 10 14 l l C D B A V l X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X XRead More

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- Chapter 37 : Magnetic Properties of Matter - HC Verma Solution, Physics
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