Page 1 35.1 CHAPTER – 35 MAGNETIC FIELD DUE TO CURRENT 1. F = B q ? ? ? ? or, B = ? q F = ? ?T F = . sec / . sec . A N = m A N ? B = r 2 0 ? ? ? or, ? 0 = ? ?rB 2 = A m A N m ? ? ? = 2 A N 2. i = 10 A, d = 1 m B = r 2 i 0 ? ? = 1 2 10 4 10 7 ? ? ? ? ? ? = 20 × 10 –6 T = 2 ?T Along +ve Y direction. ? 3. d = 1.6 mm So, r = 0.8 mm = 0.0008 m i = 20 A B ? = r 2 i 0 ? ? = 4 7 10 8 2 20 10 4 ? ? ? ? ? ? ? ? ? = 5 × 10 –3 T = 5 mT 4. i = 100 A, d = 8 m B = r 2 i 0 ? ? = 8 2 100 10 4 7 ? ? ? ? ? ? ? = 2.5 ?T ? 5. ? 0 = 4 ? × 10 –7 T-m/A r = 2 cm = 0.02 m, ? = 1 A, B ? = 1 × 10 –5 T We know: Magnetic field due to a long straight wire carrying current = r 2 0 ? ? ? B ? at P = 02 . 0 2 1 10 4 7 ? ? ? ? ? ? = 1 × 10 –5 T upward net B = 2 × 1 × 10 –7 T = 20 ?T B at Q = 1 × 10 –5 T downwards Hence net B ? = 0 ? 6. (a) The maximum magnetic field is r 2 B 0 ? ? ? ? which are along the left keeping the sense along the direction of traveling current. (b)The minimum r 2 B 0 ? ? ? ? If r = B 2 0 ? ? ? B net = 0 r < B 2 0 ? ? ? B net = 0 r > B 2 0 ? ? ? B net = r 2 B 0 ? ? ? ? 7. ? 0 = 4 ? × 10 –7 T-m/A, ? = 30 A, B = 4.0 × 10 –4 T Parallel to current. B ? due to wore at a pt. 2 cm = r 2 0 ? ? ? = 02 . 0 2 30 10 4 7 ? ? ? ? ? ? = 3 × 10 –4 T net field = ? ? ? ? 2 4 2 4 10 4 10 3 ? ? ? ? ? = 5 × 10 –4 T 1 m X axis Z axis r 8 m 100 A i ? Q P 2 cm 2 cm r i r 2 i 0 ? ? 30 A B ? = 40 × 10 –4 T – – – – – – – – – – – – – – – – – – – – Page 2 35.1 CHAPTER – 35 MAGNETIC FIELD DUE TO CURRENT 1. F = B q ? ? ? ? or, B = ? q F = ? ?T F = . sec / . sec . A N = m A N ? B = r 2 0 ? ? ? or, ? 0 = ? ?rB 2 = A m A N m ? ? ? = 2 A N 2. i = 10 A, d = 1 m B = r 2 i 0 ? ? = 1 2 10 4 10 7 ? ? ? ? ? ? = 20 × 10 –6 T = 2 ?T Along +ve Y direction. ? 3. d = 1.6 mm So, r = 0.8 mm = 0.0008 m i = 20 A B ? = r 2 i 0 ? ? = 4 7 10 8 2 20 10 4 ? ? ? ? ? ? ? ? ? = 5 × 10 –3 T = 5 mT 4. i = 100 A, d = 8 m B = r 2 i 0 ? ? = 8 2 100 10 4 7 ? ? ? ? ? ? ? = 2.5 ?T ? 5. ? 0 = 4 ? × 10 –7 T-m/A r = 2 cm = 0.02 m, ? = 1 A, B ? = 1 × 10 –5 T We know: Magnetic field due to a long straight wire carrying current = r 2 0 ? ? ? B ? at P = 02 . 0 2 1 10 4 7 ? ? ? ? ? ? = 1 × 10 –5 T upward net B = 2 × 1 × 10 –7 T = 20 ?T B at Q = 1 × 10 –5 T downwards Hence net B ? = 0 ? 6. (a) The maximum magnetic field is r 2 B 0 ? ? ? ? which are along the left keeping the sense along the direction of traveling current. (b)The minimum r 2 B 0 ? ? ? ? If r = B 2 0 ? ? ? B net = 0 r < B 2 0 ? ? ? B net = 0 r > B 2 0 ? ? ? B net = r 2 B 0 ? ? ? ? 7. ? 0 = 4 ? × 10 –7 T-m/A, ? = 30 A, B = 4.0 × 10 –4 T Parallel to current. B ? due to wore at a pt. 2 cm = r 2 0 ? ? ? = 02 . 0 2 30 10 4 7 ? ? ? ? ? ? = 3 × 10 –4 T net field = ? ? ? ? 2 4 2 4 10 4 10 3 ? ? ? ? ? = 5 × 10 –4 T 1 m X axis Z axis r 8 m 100 A i ? Q P 2 cm 2 cm r i r 2 i 0 ? ? 30 A B ? = 40 × 10 –4 T – – – – – – – – – – – – – – – – – – – – Magnetic Field due to Current 35.2 8. i = 10 A. ( K ˆ ) B = 2 × 10 –3 T South to North ( J ˆ ) To cancel the magnetic field the point should be choosen so that the net magnetic field is along - J ˆ direction. ? The point is along - i ˆ direction or along west of the wire. B = r 2 0 ? ? ? ? 2 × 10 –3 = r 2 10 10 4 7 ? ? ? ? ? ? ? r = 3 7 10 2 10 2 ? ? ? ? = 10 –3 m = 1 mm. 9. Let the tow wires be positioned at O & P R = OA, = 2 2 ) 02 . 0 ( ) 02 . 0 ( ? = 4 10 8 ? ? = 2.828 × 10 –2 m (a) B ? due to Q, at A 1 = 02 . 0 2 10 10 4 7 ? ? ? ? ? ? = 1 × 10 –4 T ( ?r towards up the line) B ? due to P, at A 1 = 06 . 0 2 10 10 4 7 ? ? ? ? ? ? = 0.33 × 10 –4 T ( ?r towards down the line) net B ? = 1 × 10 –4 – 0.33 × 10 –4 = 0.67 × 10 –4 T (b) B ? due to O at A 2 = 01 . 0 10 10 2 7 ? ? ? = 2 × 10 –4 T ?r down the line B ? due to P at A 2 = 03 . 0 10 10 2 7 ? ? ? = 0.67 × 10 –4 T ?r down the line net B ? at A 2 = 2 × 10 –4 + 0.67 × 10 –4 = 2.67 × 10 –4 T (c) B ? at A 3 due to O = 1 × 10 –4 T ?r towards down the line B ? at A 3 due to P = 1 × 10 –4 T ?r towards down the line Net B ? at A 3 = 2 × 10 –4 T (d) B ? at A 4 due to O = 2 7 10 828 . 2 10 10 2 ? ? ? ? ? = 0.7 × 10 –4 T towards SE B ? at A 4 due to P = 0.7 × 10 –4 T towards SW Net B ? = ? ? ? ? 2 4 - 2 4 - 10 0.7 10 0.7 ? ? ? = 0.989 ×10 –4 ˜ 1 × 10 –4 T 10. Cos ? = ½ , ? = 60° & ?AOB = 60° B = r 2 0 ? ? ? = 2 7 10 2 10 2 10 ? ? ? ? ? = 10 –4 T So net is [(10 –4 ) 2 + (10 –4 ) 2 + 2(10 –8 ) Cos 60°] 1/2 = 10 –4 [1 + 1 + 2 × ½ ] 1/2 = 10 -4 × 3 T = 1.732 × 10 –4 T 11. (a) B ? for X = B ? for Y Both are oppositely directed hence net B ? = 0 (b) B ? due to X = B ? due to X both directed along Z–axis Net B ? = 1 5 2 10 2 7 ? ? ? ? = 2 × 10 –6 T = 2 ?T ? (c) B ? due to X = B ? due to Y both directed opposite to each other. Hence Net B ? = 0 (d) B ? due to X = B ? due to Y = 1 × 10 –6 T both directed along (–) ve Z–axis Hence Net B ? = 2 × 1.0 × 10 –6 = 2 ?T ? A 1 ? O ? A 4 A 3 A 2 2 cm (1, 1) (–1, 1) (–1, –1) (1, –1) ? ? B A 2 cm O 2 cm 2 cm Page 3 35.1 CHAPTER – 35 MAGNETIC FIELD DUE TO CURRENT 1. F = B q ? ? ? ? or, B = ? q F = ? ?T F = . sec / . sec . A N = m A N ? B = r 2 0 ? ? ? or, ? 0 = ? ?rB 2 = A m A N m ? ? ? = 2 A N 2. i = 10 A, d = 1 m B = r 2 i 0 ? ? = 1 2 10 4 10 7 ? ? ? ? ? ? = 20 × 10 –6 T = 2 ?T Along +ve Y direction. ? 3. d = 1.6 mm So, r = 0.8 mm = 0.0008 m i = 20 A B ? = r 2 i 0 ? ? = 4 7 10 8 2 20 10 4 ? ? ? ? ? ? ? ? ? = 5 × 10 –3 T = 5 mT 4. i = 100 A, d = 8 m B = r 2 i 0 ? ? = 8 2 100 10 4 7 ? ? ? ? ? ? ? = 2.5 ?T ? 5. ? 0 = 4 ? × 10 –7 T-m/A r = 2 cm = 0.02 m, ? = 1 A, B ? = 1 × 10 –5 T We know: Magnetic field due to a long straight wire carrying current = r 2 0 ? ? ? B ? at P = 02 . 0 2 1 10 4 7 ? ? ? ? ? ? = 1 × 10 –5 T upward net B = 2 × 1 × 10 –7 T = 20 ?T B at Q = 1 × 10 –5 T downwards Hence net B ? = 0 ? 6. (a) The maximum magnetic field is r 2 B 0 ? ? ? ? which are along the left keeping the sense along the direction of traveling current. (b)The minimum r 2 B 0 ? ? ? ? If r = B 2 0 ? ? ? B net = 0 r < B 2 0 ? ? ? B net = 0 r > B 2 0 ? ? ? B net = r 2 B 0 ? ? ? ? 7. ? 0 = 4 ? × 10 –7 T-m/A, ? = 30 A, B = 4.0 × 10 –4 T Parallel to current. B ? due to wore at a pt. 2 cm = r 2 0 ? ? ? = 02 . 0 2 30 10 4 7 ? ? ? ? ? ? = 3 × 10 –4 T net field = ? ? ? ? 2 4 2 4 10 4 10 3 ? ? ? ? ? = 5 × 10 –4 T 1 m X axis Z axis r 8 m 100 A i ? Q P 2 cm 2 cm r i r 2 i 0 ? ? 30 A B ? = 40 × 10 –4 T – – – – – – – – – – – – – – – – – – – – Magnetic Field due to Current 35.2 8. i = 10 A. ( K ˆ ) B = 2 × 10 –3 T South to North ( J ˆ ) To cancel the magnetic field the point should be choosen so that the net magnetic field is along - J ˆ direction. ? The point is along - i ˆ direction or along west of the wire. B = r 2 0 ? ? ? ? 2 × 10 –3 = r 2 10 10 4 7 ? ? ? ? ? ? ? r = 3 7 10 2 10 2 ? ? ? ? = 10 –3 m = 1 mm. 9. Let the tow wires be positioned at O & P R = OA, = 2 2 ) 02 . 0 ( ) 02 . 0 ( ? = 4 10 8 ? ? = 2.828 × 10 –2 m (a) B ? due to Q, at A 1 = 02 . 0 2 10 10 4 7 ? ? ? ? ? ? = 1 × 10 –4 T ( ?r towards up the line) B ? due to P, at A 1 = 06 . 0 2 10 10 4 7 ? ? ? ? ? ? = 0.33 × 10 –4 T ( ?r towards down the line) net B ? = 1 × 10 –4 – 0.33 × 10 –4 = 0.67 × 10 –4 T (b) B ? due to O at A 2 = 01 . 0 10 10 2 7 ? ? ? = 2 × 10 –4 T ?r down the line B ? due to P at A 2 = 03 . 0 10 10 2 7 ? ? ? = 0.67 × 10 –4 T ?r down the line net B ? at A 2 = 2 × 10 –4 + 0.67 × 10 –4 = 2.67 × 10 –4 T (c) B ? at A 3 due to O = 1 × 10 –4 T ?r towards down the line B ? at A 3 due to P = 1 × 10 –4 T ?r towards down the line Net B ? at A 3 = 2 × 10 –4 T (d) B ? at A 4 due to O = 2 7 10 828 . 2 10 10 2 ? ? ? ? ? = 0.7 × 10 –4 T towards SE B ? at A 4 due to P = 0.7 × 10 –4 T towards SW Net B ? = ? ? ? ? 2 4 - 2 4 - 10 0.7 10 0.7 ? ? ? = 0.989 ×10 –4 ˜ 1 × 10 –4 T 10. Cos ? = ½ , ? = 60° & ?AOB = 60° B = r 2 0 ? ? ? = 2 7 10 2 10 2 10 ? ? ? ? ? = 10 –4 T So net is [(10 –4 ) 2 + (10 –4 ) 2 + 2(10 –8 ) Cos 60°] 1/2 = 10 –4 [1 + 1 + 2 × ½ ] 1/2 = 10 -4 × 3 T = 1.732 × 10 –4 T 11. (a) B ? for X = B ? for Y Both are oppositely directed hence net B ? = 0 (b) B ? due to X = B ? due to X both directed along Z–axis Net B ? = 1 5 2 10 2 7 ? ? ? ? = 2 × 10 –6 T = 2 ?T ? (c) B ? due to X = B ? due to Y both directed opposite to each other. Hence Net B ? = 0 (d) B ? due to X = B ? due to Y = 1 × 10 –6 T both directed along (–) ve Z–axis Hence Net B ? = 2 × 1.0 × 10 –6 = 2 ?T ? A 1 ? O ? A 4 A 3 A 2 2 cm (1, 1) (–1, 1) (–1, –1) (1, –1) ? ? B A 2 cm O 2 cm 2 cm Magnetic Field due to Current 35.3 12. (a) For each of the wire Magnitude of magnetic field = ) 45 Sin 45 Sin ( r 4 i 0 ? ? ? ? ? = ? ? 2 2 2 / 5 4 5 0 ? ? ? ? For AB ? for BC ? For CD ? and for DA ?. The two ? and 2 ? fields cancel each other. Thus B net = 0 (b) At point Q 1 due to (1) B = 2 0 10 5 . 2 2 i ? ? ? ? ? = 2 7 10 5 2 10 2 5 4 ? ? ? ? ? ? ? ? ? = 4 × 10 –5 ? due to (2) B = 2 0 10 ) 2 / 15 ( 2 i ? ? ? ? ? = 2 7 10 15 2 10 2 5 4 ? ? ? ? ? ? ? ? ? = (4/3) × 10 –5 ? due to (3) B = 2 0 10 ) 2 / 5 5 ( 2 i ? ? ? ? ? ? = 2 7 10 15 2 10 2 5 4 ? ? ? ? ? ? ? ? ? = (4/3) × 10 –5 ? due to (4) B = 2 0 10 5 . 2 2 i ? ? ? ? ? = 2 7 10 5 2 10 2 5 4 ? ? ? ? ? ? ? ? ? = 4 × 10 –5 ? B net = [4 + 4 + (4/3) + (4/3)] × 10 –5 = 3 32 × 10 –5 = 10.6 × 10 –5 ˜ 1.1 × 10 –4 T At point Q 2 due to (1) 2 o 10 ) 5 . 2 ( 2 i ? ? ? ? ? ? due to (2) 2 o 10 ) 2 / 15 ( 2 i ? ? ? ? ? ? due to (3) 2 o 10 ) 5 . 2 ( 2 i ? ? ? ? ? ? due to (4) 2 o 10 ) 2 / 15 ( 2 i ? ? ? ? ? ? B net = 0 At point Q 3 due to (1) 2 7 10 ) 2 / 15 ( 2 5 10 4 ? ? ? ? ? ? ? ? = 4/3 × 10 –5 ? due to (2) 2 7 10 ) 2 / 5 ( 2 5 10 4 ? ? ? ? ? ? ? ? = 4 × 10 –5 ? due to (3) 2 7 10 ) 2 / 5 ( 2 5 10 4 ? ? ? ? ? ? ? ? = 4 × 10 –5 ? due to (4) 2 7 10 ) 2 / 15 ( 2 5 10 4 ? ? ? ? ? ? ? ? = 4/3 × 10 –5 ? B net = [4 + 4 + (4/3) + (4/3)] × 10 –5 = 3 32 × 10 –5 = 10.6 × 10 –5 ˜ 1.1 × 10 –4 T For Q 4 due to (1) 4/3 × 10 –5 ? due to (2) 4 × 10 –5 ? due to (3) 4/3 × 10 –5 ? due to (4) 4 × 10 –5 ? B net = 0 P D C 4 3 B A 5 cm 2 5 2 / 2 5 Q 1 Q 2 Q 3 Q 4 Page 4 35.1 CHAPTER – 35 MAGNETIC FIELD DUE TO CURRENT 1. F = B q ? ? ? ? or, B = ? q F = ? ?T F = . sec / . sec . A N = m A N ? B = r 2 0 ? ? ? or, ? 0 = ? ?rB 2 = A m A N m ? ? ? = 2 A N 2. i = 10 A, d = 1 m B = r 2 i 0 ? ? = 1 2 10 4 10 7 ? ? ? ? ? ? = 20 × 10 –6 T = 2 ?T Along +ve Y direction. ? 3. d = 1.6 mm So, r = 0.8 mm = 0.0008 m i = 20 A B ? = r 2 i 0 ? ? = 4 7 10 8 2 20 10 4 ? ? ? ? ? ? ? ? ? = 5 × 10 –3 T = 5 mT 4. i = 100 A, d = 8 m B = r 2 i 0 ? ? = 8 2 100 10 4 7 ? ? ? ? ? ? ? = 2.5 ?T ? 5. ? 0 = 4 ? × 10 –7 T-m/A r = 2 cm = 0.02 m, ? = 1 A, B ? = 1 × 10 –5 T We know: Magnetic field due to a long straight wire carrying current = r 2 0 ? ? ? B ? at P = 02 . 0 2 1 10 4 7 ? ? ? ? ? ? = 1 × 10 –5 T upward net B = 2 × 1 × 10 –7 T = 20 ?T B at Q = 1 × 10 –5 T downwards Hence net B ? = 0 ? 6. (a) The maximum magnetic field is r 2 B 0 ? ? ? ? which are along the left keeping the sense along the direction of traveling current. (b)The minimum r 2 B 0 ? ? ? ? If r = B 2 0 ? ? ? B net = 0 r < B 2 0 ? ? ? B net = 0 r > B 2 0 ? ? ? B net = r 2 B 0 ? ? ? ? 7. ? 0 = 4 ? × 10 –7 T-m/A, ? = 30 A, B = 4.0 × 10 –4 T Parallel to current. B ? due to wore at a pt. 2 cm = r 2 0 ? ? ? = 02 . 0 2 30 10 4 7 ? ? ? ? ? ? = 3 × 10 –4 T net field = ? ? ? ? 2 4 2 4 10 4 10 3 ? ? ? ? ? = 5 × 10 –4 T 1 m X axis Z axis r 8 m 100 A i ? Q P 2 cm 2 cm r i r 2 i 0 ? ? 30 A B ? = 40 × 10 –4 T – – – – – – – – – – – – – – – – – – – – Magnetic Field due to Current 35.2 8. i = 10 A. ( K ˆ ) B = 2 × 10 –3 T South to North ( J ˆ ) To cancel the magnetic field the point should be choosen so that the net magnetic field is along - J ˆ direction. ? The point is along - i ˆ direction or along west of the wire. B = r 2 0 ? ? ? ? 2 × 10 –3 = r 2 10 10 4 7 ? ? ? ? ? ? ? r = 3 7 10 2 10 2 ? ? ? ? = 10 –3 m = 1 mm. 9. Let the tow wires be positioned at O & P R = OA, = 2 2 ) 02 . 0 ( ) 02 . 0 ( ? = 4 10 8 ? ? = 2.828 × 10 –2 m (a) B ? due to Q, at A 1 = 02 . 0 2 10 10 4 7 ? ? ? ? ? ? = 1 × 10 –4 T ( ?r towards up the line) B ? due to P, at A 1 = 06 . 0 2 10 10 4 7 ? ? ? ? ? ? = 0.33 × 10 –4 T ( ?r towards down the line) net B ? = 1 × 10 –4 – 0.33 × 10 –4 = 0.67 × 10 –4 T (b) B ? due to O at A 2 = 01 . 0 10 10 2 7 ? ? ? = 2 × 10 –4 T ?r down the line B ? due to P at A 2 = 03 . 0 10 10 2 7 ? ? ? = 0.67 × 10 –4 T ?r down the line net B ? at A 2 = 2 × 10 –4 + 0.67 × 10 –4 = 2.67 × 10 –4 T (c) B ? at A 3 due to O = 1 × 10 –4 T ?r towards down the line B ? at A 3 due to P = 1 × 10 –4 T ?r towards down the line Net B ? at A 3 = 2 × 10 –4 T (d) B ? at A 4 due to O = 2 7 10 828 . 2 10 10 2 ? ? ? ? ? = 0.7 × 10 –4 T towards SE B ? at A 4 due to P = 0.7 × 10 –4 T towards SW Net B ? = ? ? ? ? 2 4 - 2 4 - 10 0.7 10 0.7 ? ? ? = 0.989 ×10 –4 ˜ 1 × 10 –4 T 10. Cos ? = ½ , ? = 60° & ?AOB = 60° B = r 2 0 ? ? ? = 2 7 10 2 10 2 10 ? ? ? ? ? = 10 –4 T So net is [(10 –4 ) 2 + (10 –4 ) 2 + 2(10 –8 ) Cos 60°] 1/2 = 10 –4 [1 + 1 + 2 × ½ ] 1/2 = 10 -4 × 3 T = 1.732 × 10 –4 T 11. (a) B ? for X = B ? for Y Both are oppositely directed hence net B ? = 0 (b) B ? due to X = B ? due to X both directed along Z–axis Net B ? = 1 5 2 10 2 7 ? ? ? ? = 2 × 10 –6 T = 2 ?T ? (c) B ? due to X = B ? due to Y both directed opposite to each other. Hence Net B ? = 0 (d) B ? due to X = B ? due to Y = 1 × 10 –6 T both directed along (–) ve Z–axis Hence Net B ? = 2 × 1.0 × 10 –6 = 2 ?T ? A 1 ? O ? A 4 A 3 A 2 2 cm (1, 1) (–1, 1) (–1, –1) (1, –1) ? ? B A 2 cm O 2 cm 2 cm Magnetic Field due to Current 35.3 12. (a) For each of the wire Magnitude of magnetic field = ) 45 Sin 45 Sin ( r 4 i 0 ? ? ? ? ? = ? ? 2 2 2 / 5 4 5 0 ? ? ? ? For AB ? for BC ? For CD ? and for DA ?. The two ? and 2 ? fields cancel each other. Thus B net = 0 (b) At point Q 1 due to (1) B = 2 0 10 5 . 2 2 i ? ? ? ? ? = 2 7 10 5 2 10 2 5 4 ? ? ? ? ? ? ? ? ? = 4 × 10 –5 ? due to (2) B = 2 0 10 ) 2 / 15 ( 2 i ? ? ? ? ? = 2 7 10 15 2 10 2 5 4 ? ? ? ? ? ? ? ? ? = (4/3) × 10 –5 ? due to (3) B = 2 0 10 ) 2 / 5 5 ( 2 i ? ? ? ? ? ? = 2 7 10 15 2 10 2 5 4 ? ? ? ? ? ? ? ? ? = (4/3) × 10 –5 ? due to (4) B = 2 0 10 5 . 2 2 i ? ? ? ? ? = 2 7 10 5 2 10 2 5 4 ? ? ? ? ? ? ? ? ? = 4 × 10 –5 ? B net = [4 + 4 + (4/3) + (4/3)] × 10 –5 = 3 32 × 10 –5 = 10.6 × 10 –5 ˜ 1.1 × 10 –4 T At point Q 2 due to (1) 2 o 10 ) 5 . 2 ( 2 i ? ? ? ? ? ? due to (2) 2 o 10 ) 2 / 15 ( 2 i ? ? ? ? ? ? due to (3) 2 o 10 ) 5 . 2 ( 2 i ? ? ? ? ? ? due to (4) 2 o 10 ) 2 / 15 ( 2 i ? ? ? ? ? ? B net = 0 At point Q 3 due to (1) 2 7 10 ) 2 / 15 ( 2 5 10 4 ? ? ? ? ? ? ? ? = 4/3 × 10 –5 ? due to (2) 2 7 10 ) 2 / 5 ( 2 5 10 4 ? ? ? ? ? ? ? ? = 4 × 10 –5 ? due to (3) 2 7 10 ) 2 / 5 ( 2 5 10 4 ? ? ? ? ? ? ? ? = 4 × 10 –5 ? due to (4) 2 7 10 ) 2 / 15 ( 2 5 10 4 ? ? ? ? ? ? ? ? = 4/3 × 10 –5 ? B net = [4 + 4 + (4/3) + (4/3)] × 10 –5 = 3 32 × 10 –5 = 10.6 × 10 –5 ˜ 1.1 × 10 –4 T For Q 4 due to (1) 4/3 × 10 –5 ? due to (2) 4 × 10 –5 ? due to (3) 4/3 × 10 –5 ? due to (4) 4 × 10 –5 ? B net = 0 P D C 4 3 B A 5 cm 2 5 2 / 2 5 Q 1 Q 2 Q 3 Q 4 Magnetic Field due to Current 35.4 13. Since all the points lie along a circle with radius = ‘d’ Hence ‘R’ & ‘Q’ both at a distance ‘d’ from the wire. So, magnetic field B ? due to are same in magnitude. As the wires can be treated as semi infinite straight current carrying conductors. Hence magnetic field B ? = d 4 i 0 ? ? At P B 1 due to 1 is 0 B 2 due to 2 is d 4 i 0 ? ? At Q B 1 due to 1 is d 4 i 0 ? ? B 2 due to 2 is 0 At R B 1 due to 1 is 0 B 2 due to 2 is d 4 i 0 ? ? At S B 1 due to 1 is d 4 i 0 ? ? B 2 due to 2 is 0 14. B = d 4 i 0 ? ? 2 Sin ? = 4 x d 2 x 2 d 4 i 2 2 0 ? ? ? ? ? = 4 x d d 4 ix 2 2 0 ? ? ? (a) When d >> x Neglecting x w.r.t. d B = 2 0 d d ix ?? ? = 2 0 d ix ?? ? ? B ? 2 d 1 (b) When x >> d, neglecting d w.r.t. x B = 2 / dx 4 ix 0 ? ? = d 4 i 2 0 ? ? ? B ? d 1 15. ? = 10 A, a = 10 cm = 0.1 m r = OP = 1 . 0 2 3 ? m B = ) Sin Sin ( r 4 2 1 0 ? ? ? ? ? ? = 1 . 0 2 3 1 10 10 7 ? ? ? ? = 732 . 1 10 2 5 ? ? = 1.154 × 10 –5 T = 11.54 ?T P i 1 i Q R S d 2 d i ? ? ? ? x x/2 O 10 A P B A 10 cm Q 1 Q 2 30 30 P Page 5 35.1 CHAPTER – 35 MAGNETIC FIELD DUE TO CURRENT 1. F = B q ? ? ? ? or, B = ? q F = ? ?T F = . sec / . sec . A N = m A N ? B = r 2 0 ? ? ? or, ? 0 = ? ?rB 2 = A m A N m ? ? ? = 2 A N 2. i = 10 A, d = 1 m B = r 2 i 0 ? ? = 1 2 10 4 10 7 ? ? ? ? ? ? = 20 × 10 –6 T = 2 ?T Along +ve Y direction. ? 3. d = 1.6 mm So, r = 0.8 mm = 0.0008 m i = 20 A B ? = r 2 i 0 ? ? = 4 7 10 8 2 20 10 4 ? ? ? ? ? ? ? ? ? = 5 × 10 –3 T = 5 mT 4. i = 100 A, d = 8 m B = r 2 i 0 ? ? = 8 2 100 10 4 7 ? ? ? ? ? ? ? = 2.5 ?T ? 5. ? 0 = 4 ? × 10 –7 T-m/A r = 2 cm = 0.02 m, ? = 1 A, B ? = 1 × 10 –5 T We know: Magnetic field due to a long straight wire carrying current = r 2 0 ? ? ? B ? at P = 02 . 0 2 1 10 4 7 ? ? ? ? ? ? = 1 × 10 –5 T upward net B = 2 × 1 × 10 –7 T = 20 ?T B at Q = 1 × 10 –5 T downwards Hence net B ? = 0 ? 6. (a) The maximum magnetic field is r 2 B 0 ? ? ? ? which are along the left keeping the sense along the direction of traveling current. (b)The minimum r 2 B 0 ? ? ? ? If r = B 2 0 ? ? ? B net = 0 r < B 2 0 ? ? ? B net = 0 r > B 2 0 ? ? ? B net = r 2 B 0 ? ? ? ? 7. ? 0 = 4 ? × 10 –7 T-m/A, ? = 30 A, B = 4.0 × 10 –4 T Parallel to current. B ? due to wore at a pt. 2 cm = r 2 0 ? ? ? = 02 . 0 2 30 10 4 7 ? ? ? ? ? ? = 3 × 10 –4 T net field = ? ? ? ? 2 4 2 4 10 4 10 3 ? ? ? ? ? = 5 × 10 –4 T 1 m X axis Z axis r 8 m 100 A i ? Q P 2 cm 2 cm r i r 2 i 0 ? ? 30 A B ? = 40 × 10 –4 T – – – – – – – – – – – – – – – – – – – – Magnetic Field due to Current 35.2 8. i = 10 A. ( K ˆ ) B = 2 × 10 –3 T South to North ( J ˆ ) To cancel the magnetic field the point should be choosen so that the net magnetic field is along - J ˆ direction. ? The point is along - i ˆ direction or along west of the wire. B = r 2 0 ? ? ? ? 2 × 10 –3 = r 2 10 10 4 7 ? ? ? ? ? ? ? r = 3 7 10 2 10 2 ? ? ? ? = 10 –3 m = 1 mm. 9. Let the tow wires be positioned at O & P R = OA, = 2 2 ) 02 . 0 ( ) 02 . 0 ( ? = 4 10 8 ? ? = 2.828 × 10 –2 m (a) B ? due to Q, at A 1 = 02 . 0 2 10 10 4 7 ? ? ? ? ? ? = 1 × 10 –4 T ( ?r towards up the line) B ? due to P, at A 1 = 06 . 0 2 10 10 4 7 ? ? ? ? ? ? = 0.33 × 10 –4 T ( ?r towards down the line) net B ? = 1 × 10 –4 – 0.33 × 10 –4 = 0.67 × 10 –4 T (b) B ? due to O at A 2 = 01 . 0 10 10 2 7 ? ? ? = 2 × 10 –4 T ?r down the line B ? due to P at A 2 = 03 . 0 10 10 2 7 ? ? ? = 0.67 × 10 –4 T ?r down the line net B ? at A 2 = 2 × 10 –4 + 0.67 × 10 –4 = 2.67 × 10 –4 T (c) B ? at A 3 due to O = 1 × 10 –4 T ?r towards down the line B ? at A 3 due to P = 1 × 10 –4 T ?r towards down the line Net B ? at A 3 = 2 × 10 –4 T (d) B ? at A 4 due to O = 2 7 10 828 . 2 10 10 2 ? ? ? ? ? = 0.7 × 10 –4 T towards SE B ? at A 4 due to P = 0.7 × 10 –4 T towards SW Net B ? = ? ? ? ? 2 4 - 2 4 - 10 0.7 10 0.7 ? ? ? = 0.989 ×10 –4 ˜ 1 × 10 –4 T 10. Cos ? = ½ , ? = 60° & ?AOB = 60° B = r 2 0 ? ? ? = 2 7 10 2 10 2 10 ? ? ? ? ? = 10 –4 T So net is [(10 –4 ) 2 + (10 –4 ) 2 + 2(10 –8 ) Cos 60°] 1/2 = 10 –4 [1 + 1 + 2 × ½ ] 1/2 = 10 -4 × 3 T = 1.732 × 10 –4 T 11. (a) B ? for X = B ? for Y Both are oppositely directed hence net B ? = 0 (b) B ? due to X = B ? due to X both directed along Z–axis Net B ? = 1 5 2 10 2 7 ? ? ? ? = 2 × 10 –6 T = 2 ?T ? (c) B ? due to X = B ? due to Y both directed opposite to each other. Hence Net B ? = 0 (d) B ? due to X = B ? due to Y = 1 × 10 –6 T both directed along (–) ve Z–axis Hence Net B ? = 2 × 1.0 × 10 –6 = 2 ?T ? A 1 ? O ? A 4 A 3 A 2 2 cm (1, 1) (–1, 1) (–1, –1) (1, –1) ? ? B A 2 cm O 2 cm 2 cm Magnetic Field due to Current 35.3 12. (a) For each of the wire Magnitude of magnetic field = ) 45 Sin 45 Sin ( r 4 i 0 ? ? ? ? ? = ? ? 2 2 2 / 5 4 5 0 ? ? ? ? For AB ? for BC ? For CD ? and for DA ?. The two ? and 2 ? fields cancel each other. Thus B net = 0 (b) At point Q 1 due to (1) B = 2 0 10 5 . 2 2 i ? ? ? ? ? = 2 7 10 5 2 10 2 5 4 ? ? ? ? ? ? ? ? ? = 4 × 10 –5 ? due to (2) B = 2 0 10 ) 2 / 15 ( 2 i ? ? ? ? ? = 2 7 10 15 2 10 2 5 4 ? ? ? ? ? ? ? ? ? = (4/3) × 10 –5 ? due to (3) B = 2 0 10 ) 2 / 5 5 ( 2 i ? ? ? ? ? ? = 2 7 10 15 2 10 2 5 4 ? ? ? ? ? ? ? ? ? = (4/3) × 10 –5 ? due to (4) B = 2 0 10 5 . 2 2 i ? ? ? ? ? = 2 7 10 5 2 10 2 5 4 ? ? ? ? ? ? ? ? ? = 4 × 10 –5 ? B net = [4 + 4 + (4/3) + (4/3)] × 10 –5 = 3 32 × 10 –5 = 10.6 × 10 –5 ˜ 1.1 × 10 –4 T At point Q 2 due to (1) 2 o 10 ) 5 . 2 ( 2 i ? ? ? ? ? ? due to (2) 2 o 10 ) 2 / 15 ( 2 i ? ? ? ? ? ? due to (3) 2 o 10 ) 5 . 2 ( 2 i ? ? ? ? ? ? due to (4) 2 o 10 ) 2 / 15 ( 2 i ? ? ? ? ? ? B net = 0 At point Q 3 due to (1) 2 7 10 ) 2 / 15 ( 2 5 10 4 ? ? ? ? ? ? ? ? = 4/3 × 10 –5 ? due to (2) 2 7 10 ) 2 / 5 ( 2 5 10 4 ? ? ? ? ? ? ? ? = 4 × 10 –5 ? due to (3) 2 7 10 ) 2 / 5 ( 2 5 10 4 ? ? ? ? ? ? ? ? = 4 × 10 –5 ? due to (4) 2 7 10 ) 2 / 15 ( 2 5 10 4 ? ? ? ? ? ? ? ? = 4/3 × 10 –5 ? B net = [4 + 4 + (4/3) + (4/3)] × 10 –5 = 3 32 × 10 –5 = 10.6 × 10 –5 ˜ 1.1 × 10 –4 T For Q 4 due to (1) 4/3 × 10 –5 ? due to (2) 4 × 10 –5 ? due to (3) 4/3 × 10 –5 ? due to (4) 4 × 10 –5 ? B net = 0 P D C 4 3 B A 5 cm 2 5 2 / 2 5 Q 1 Q 2 Q 3 Q 4 Magnetic Field due to Current 35.4 13. Since all the points lie along a circle with radius = ‘d’ Hence ‘R’ & ‘Q’ both at a distance ‘d’ from the wire. So, magnetic field B ? due to are same in magnitude. As the wires can be treated as semi infinite straight current carrying conductors. Hence magnetic field B ? = d 4 i 0 ? ? At P B 1 due to 1 is 0 B 2 due to 2 is d 4 i 0 ? ? At Q B 1 due to 1 is d 4 i 0 ? ? B 2 due to 2 is 0 At R B 1 due to 1 is 0 B 2 due to 2 is d 4 i 0 ? ? At S B 1 due to 1 is d 4 i 0 ? ? B 2 due to 2 is 0 14. B = d 4 i 0 ? ? 2 Sin ? = 4 x d 2 x 2 d 4 i 2 2 0 ? ? ? ? ? = 4 x d d 4 ix 2 2 0 ? ? ? (a) When d >> x Neglecting x w.r.t. d B = 2 0 d d ix ?? ? = 2 0 d ix ?? ? ? B ? 2 d 1 (b) When x >> d, neglecting d w.r.t. x B = 2 / dx 4 ix 0 ? ? = d 4 i 2 0 ? ? ? B ? d 1 15. ? = 10 A, a = 10 cm = 0.1 m r = OP = 1 . 0 2 3 ? m B = ) Sin Sin ( r 4 2 1 0 ? ? ? ? ? ? = 1 . 0 2 3 1 10 10 7 ? ? ? ? = 732 . 1 10 2 5 ? ? = 1.154 × 10 –5 T = 11.54 ?T P i 1 i Q R S d 2 d i ? ? ? ? x x/2 O 10 A P B A 10 cm Q 1 Q 2 30 30 P Magnetic Field due to Current 35.5 16. B 1 = d 2 i 0 ? ? , B 2 = ) Sin 2 ( d 4 i 0 ? ? ? ? = 4 d 2 2 d 4 i 2 2 0 ? ? ? ? ? ? = 4 d d 4 i 2 2 0 ? ? ? ? ? B 1 – B 2 = 100 1 B 2 ? 4 d d 4 i d 2 i 2 2 0 0 ? ? ? ? ? ? ? ? = d 200 i 0 ? ? ? 4 d d 4 i 2 2 0 ? ? ? ? ? = ? ? ? ? ? ? ? ? ? 200 1 2 1 d i 0 ? 4 d 4 2 2 ? ? ? = 200 99 ? 4 d 2 2 2 ? ? ? = 2 200 4 99 ? ? ? ? ? ? ? = 40000 156816 = 3.92 ? l 2 = 3.92 d 2 + 2 4 92 . 3 ? 2 4 92 . 3 1 ? ? ? ? ? ? ? ? = 3.92 d 2 ? 0.02 l 2 = 3.92 d 2 ? 2 2 d ? = 92 . 3 02 . 0 = ? d = 92 . 3 02 . 0 = 0.07 17. As resistances vary as r & 2r Hence Current along ABC = 3 i & along ADC = i 3 2 Now, B ? due to ADC = ? ? ? ? ? ? ? ? ? ? ? ? ? a 3 4 2 2 2 i 2 0 = a 3 i 2 2 0 ? ? B ? due to ABC = ? ? ? ? ? ? ? ? ? ? ? ? a 3 4 2 2 i 2 0 = a 6 i 2 2 0 ? ? Now B ? = a 3 i 2 2 0 ? ? – a 6 i 2 2 0 ? ? = a 3 i 2 0 ? ? ? 18. A 0 = 4 a 16 a 2 2 ? = 16 a 5 2 = 4 5 a D 0 = 2 2 2 a 4 a 3 ? ? ? ? ? ? ? ? ? ? ? ? ? = 4 a 16 a 9 2 2 ? = 16 a 13 2 = 4 13 a Magnetic field due to AB B AB = ? ? 4 / a 2 i 4 0 ? ? ? (Sin (90 – i) + Sin (90 – ?)) = ? ? ? ? Cos 2 a 4 i 2 0 = ? ? ) 4 / 5 ( a 2 / a 2 a 4 i 2 0 ? ? ? ? ? = 5 i 2 0 ? ? Magnetic field due to DC B DC = ? ? 4 / a 3 2 i 4 0 ? ? ? 2Sin (90° – B) = ? ? ? ? ? ? Cos a 3 4 2 4 i 0 = ? ? ) 4 / a 13 ( 2 / a a 3 i 0 ? ? ? ? = 13 3 a i 2 0 ? ? The magnetic field due to AD & BC are equal and appropriate hence cancle each other. Hence, net magnetic field is 5 i 2 0 ? ? – 13 3 a i 2 0 ? ? = ? ? ? ? ? ? ? ? ? 13 3 1 5 1 a i 2 0 d i ? ? l C D B A i/3 2i/3a a/2 2 a i D C B A i i 3a/4 a/4 O a a/2 a/2 ? ? ? ? ?? ? ?Read More

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- HC Verma Solutions: Permanent Magnets (Chapter 36)
- HC Verma Solutions: Magnetic Properties of Matter (Chapter 37)
- HC Verma Solutions: Electromagnetic Induction (Chapter 38)
- HC Verma Solutions: Alternating Current (Chapter 39)
- HC Verma Solutions: Electromagnetic Waves (Chapter 40)
- HC Verma Solutions: Electric Current through Gases (Chapter 41)