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Page 1 36.1 CHAPTER – 36 PERMANENT MAGNETS 1. m = 10 Am, d = 5 cm = 0.05 m B = 2 0 r m 4 ? ? = ? ? 2 2 7 10 5 10 10 ? ? ? ? = 25 10 2 ? = 4 × 10 –4 Tesla 2. m 1 =m 2 = 10 Am r = 2 cm = 0.02 m we know Force exerted by tow magnetic poles on each other = 2 2 1 0 r m m 4 ? ? = 4 2 7 10 4 4 10 10 4 ? ? ? ? ? ? ? ? = 2.5 × 10 –2 N 3. B = – ? d dv ? dv = –B dl = – 0.2 × 10 –3 × 0.5 = – 0.1 × 10 –3 Tm Since the sigh is –ve therefore potential decreases. 4. Here dx = 10 sin 30° cm = 5 cm dx dV = B = m 10 5 m T 10 1 . 0 2 4 ? ? ? ? ? Since B is perpendicular to equipotential surface. Here it is at angle 120° with (+ve) xaxis and B = 2 × 10 –4 T 5. B = 2 × 10 –4 T d = 10 cm = 0.1 m (a) if the point at endon postion. B = 3 0 d M 2 4 ? ? ? 2 × 10 –4 = 3 1 7 ) 10 ( M 2 10 ? ? ? ? 2 10 10 10 2 7 3 4 ? ? ? ? ? ? = M ? M = 1 Am 2 (b) If the point is at broadon position 3 0 d M 4 ? ? ? 2 × 10 –4 = 3 1 7 ) 10 ( M 10 ? ? ? ? M = 2 Am 2 6. Given : ? = tan –1 2 ? tan ? = 2 ? 2 = tan 2 ? ? tan ? = 2 cot ? ? 2 tan ? = cot ? We know 2 tan ? = tan ? Comparing we get, tan ? = cot ? or, tan ? = tan(90 – ?) or ? = 90 – ? or ? + ? = 90 Hence magnetic field due to the dipole is ?r to the magnetic axis. ? 7. Magnetic field at the broad side on position : B = ? ? 2 / 3 2 2 0 d M 4 ? ? ? ? 2l = 8 cm d = 3 cm ? 4 × 10 –6 = ? ? 2 / 3 4 4 2 7 10 16 10 9 10 8 m 10 ? ? ? ? ? ? ? ? ? ? ? 4 × 10 –6 = ? ? ? ? 2 / 3 2 / 3 4 9 25 10 8 m 10 ? ? ? ? ? ? m = 9 8 6 10 8 10 125 10 4 ? ? ? ? ? ? ? = 62.5 × 10 –5 Am N S In cm 30° X V 0.1×10 –4 Tm 30° 0.2×10 –4 Tm 0.4×10 –4 Tm 0.3×10 –4 Tm ? ? ? ? P S N ? ? Page 2 36.1 CHAPTER – 36 PERMANENT MAGNETS 1. m = 10 Am, d = 5 cm = 0.05 m B = 2 0 r m 4 ? ? = ? ? 2 2 7 10 5 10 10 ? ? ? ? = 25 10 2 ? = 4 × 10 –4 Tesla 2. m 1 =m 2 = 10 Am r = 2 cm = 0.02 m we know Force exerted by tow magnetic poles on each other = 2 2 1 0 r m m 4 ? ? = 4 2 7 10 4 4 10 10 4 ? ? ? ? ? ? ? ? = 2.5 × 10 –2 N 3. B = – ? d dv ? dv = –B dl = – 0.2 × 10 –3 × 0.5 = – 0.1 × 10 –3 Tm Since the sigh is –ve therefore potential decreases. 4. Here dx = 10 sin 30° cm = 5 cm dx dV = B = m 10 5 m T 10 1 . 0 2 4 ? ? ? ? ? Since B is perpendicular to equipotential surface. Here it is at angle 120° with (+ve) xaxis and B = 2 × 10 –4 T 5. B = 2 × 10 –4 T d = 10 cm = 0.1 m (a) if the point at endon postion. B = 3 0 d M 2 4 ? ? ? 2 × 10 –4 = 3 1 7 ) 10 ( M 2 10 ? ? ? ? 2 10 10 10 2 7 3 4 ? ? ? ? ? ? = M ? M = 1 Am 2 (b) If the point is at broadon position 3 0 d M 4 ? ? ? 2 × 10 –4 = 3 1 7 ) 10 ( M 10 ? ? ? ? M = 2 Am 2 6. Given : ? = tan –1 2 ? tan ? = 2 ? 2 = tan 2 ? ? tan ? = 2 cot ? ? 2 tan ? = cot ? We know 2 tan ? = tan ? Comparing we get, tan ? = cot ? or, tan ? = tan(90 – ?) or ? = 90 – ? or ? + ? = 90 Hence magnetic field due to the dipole is ?r to the magnetic axis. ? 7. Magnetic field at the broad side on position : B = ? ? 2 / 3 2 2 0 d M 4 ? ? ? ? 2l = 8 cm d = 3 cm ? 4 × 10 –6 = ? ? 2 / 3 4 4 2 7 10 16 10 9 10 8 m 10 ? ? ? ? ? ? ? ? ? ? ? 4 × 10 –6 = ? ? ? ? 2 / 3 2 / 3 4 9 25 10 8 m 10 ? ? ? ? ? ? m = 9 8 6 10 8 10 125 10 4 ? ? ? ? ? ? ? = 62.5 × 10 –5 Am N S In cm 30° X V 0.1×10 –4 Tm 30° 0.2×10 –4 Tm 0.4×10 –4 Tm 0.3×10 –4 Tm ? ? ? ? P S N ? ? Permanent Magnets 36.2 8. We know for a magnetic dipole with its north pointing the north, the neutral point in the broadside on position. Again B ? in this case = 3 0 d 4 M ? ? ? 3 0 d 4 M ? ? = H B due to earth ? 3 7 d 44 . 1 10 ? ? = 18 ?T ? 3 7 d 44 . 1 10 ? ? = 18 × 10 –6 ? d 3 = 8 × 10 –3 ? d = 2 × 10 –1 m = 20 cm In the plane bisecting the dipole. 9. When the magnet is such that its North faces the geographic south of earth. The neutral point lies along the axial line of the magnet. 3 0 d M 2 4 ? ? = 18 × 10 –6 ? 3 7 d 72 . 0 2 10 ? ? ? = 18 × 10 –6 ? d 3 = 10 18 10 7 . 0 2 6  7 ? ? ? ? ? d = 3 / 1 6 9 10 10 8 ? ? ? ? ? ? ? ? ? ? ? = 2 × 10 –1 m = 20 cm 10. Magnetic moment = 0.72 2 Am 2 = M B = 3 0 d M 4 ? ? B H = 18 ?T ? 3 7 d 4 2 72 . 0 10 4 ? ? ? ? ? ? = 18 × 10 –6 ? d 3 = 6 7 10 18 10 414 . 1 72 . 0 ? ? ? ? ? = 0.005656 ? d ˜ 0.2 m = 20 cm 11. The geomagnetic pole is at the end on position of the earth. B = 3 0 d M 2 4 ? ? = 3 3 22 7 ) 10 6400 ( 10 8 2 10 ? ? ? ? ? ˜ 60 × 10 –6 T = 60 ?T 12. B ? = 3.4 × 10 –5 T Given 3 0 R M 4 ? ? = 3.4 × 10 –5 ? M = 7 3 5 10 4 4 R 10 4 . 3 ? ? ? ? ? ? ? ? = 3.4 × 10 2 R 3 B ? at Poles = 3 0 R M 2 4 ? ? = = 6.8 × 10 –5 T 13. ?(dip) = 60° B H = B cos 60° ? B = 52 × 10 –6 = 52 ?T B V = B sin ? = 52 × 10 –6 2 3 = 44.98 ?T ˜ 45 ?T ? 14. If ? 1 and ? 2 be the apparent dips shown by the dip circle in the 2 ?r positions, the true dip ? is given by Cot 2 ? = Cot 2 ? 1 + Cot 2 ? 2 ? Cot 2 ? = Cot 2 45° + Cot 2 53° ? Cot 2 ? = 1.56 ? ? = 38.6 ˜ 39° ? N S P d B M S N B H N W N S S E tan –1 2 N 2 N 1 B H d Page 3 36.1 CHAPTER – 36 PERMANENT MAGNETS 1. m = 10 Am, d = 5 cm = 0.05 m B = 2 0 r m 4 ? ? = ? ? 2 2 7 10 5 10 10 ? ? ? ? = 25 10 2 ? = 4 × 10 –4 Tesla 2. m 1 =m 2 = 10 Am r = 2 cm = 0.02 m we know Force exerted by tow magnetic poles on each other = 2 2 1 0 r m m 4 ? ? = 4 2 7 10 4 4 10 10 4 ? ? ? ? ? ? ? ? = 2.5 × 10 –2 N 3. B = – ? d dv ? dv = –B dl = – 0.2 × 10 –3 × 0.5 = – 0.1 × 10 –3 Tm Since the sigh is –ve therefore potential decreases. 4. Here dx = 10 sin 30° cm = 5 cm dx dV = B = m 10 5 m T 10 1 . 0 2 4 ? ? ? ? ? Since B is perpendicular to equipotential surface. Here it is at angle 120° with (+ve) xaxis and B = 2 × 10 –4 T 5. B = 2 × 10 –4 T d = 10 cm = 0.1 m (a) if the point at endon postion. B = 3 0 d M 2 4 ? ? ? 2 × 10 –4 = 3 1 7 ) 10 ( M 2 10 ? ? ? ? 2 10 10 10 2 7 3 4 ? ? ? ? ? ? = M ? M = 1 Am 2 (b) If the point is at broadon position 3 0 d M 4 ? ? ? 2 × 10 –4 = 3 1 7 ) 10 ( M 10 ? ? ? ? M = 2 Am 2 6. Given : ? = tan –1 2 ? tan ? = 2 ? 2 = tan 2 ? ? tan ? = 2 cot ? ? 2 tan ? = cot ? We know 2 tan ? = tan ? Comparing we get, tan ? = cot ? or, tan ? = tan(90 – ?) or ? = 90 – ? or ? + ? = 90 Hence magnetic field due to the dipole is ?r to the magnetic axis. ? 7. Magnetic field at the broad side on position : B = ? ? 2 / 3 2 2 0 d M 4 ? ? ? ? 2l = 8 cm d = 3 cm ? 4 × 10 –6 = ? ? 2 / 3 4 4 2 7 10 16 10 9 10 8 m 10 ? ? ? ? ? ? ? ? ? ? ? 4 × 10 –6 = ? ? ? ? 2 / 3 2 / 3 4 9 25 10 8 m 10 ? ? ? ? ? ? m = 9 8 6 10 8 10 125 10 4 ? ? ? ? ? ? ? = 62.5 × 10 –5 Am N S In cm 30° X V 0.1×10 –4 Tm 30° 0.2×10 –4 Tm 0.4×10 –4 Tm 0.3×10 –4 Tm ? ? ? ? P S N ? ? Permanent Magnets 36.2 8. We know for a magnetic dipole with its north pointing the north, the neutral point in the broadside on position. Again B ? in this case = 3 0 d 4 M ? ? ? 3 0 d 4 M ? ? = H B due to earth ? 3 7 d 44 . 1 10 ? ? = 18 ?T ? 3 7 d 44 . 1 10 ? ? = 18 × 10 –6 ? d 3 = 8 × 10 –3 ? d = 2 × 10 –1 m = 20 cm In the plane bisecting the dipole. 9. When the magnet is such that its North faces the geographic south of earth. The neutral point lies along the axial line of the magnet. 3 0 d M 2 4 ? ? = 18 × 10 –6 ? 3 7 d 72 . 0 2 10 ? ? ? = 18 × 10 –6 ? d 3 = 10 18 10 7 . 0 2 6  7 ? ? ? ? ? d = 3 / 1 6 9 10 10 8 ? ? ? ? ? ? ? ? ? ? ? = 2 × 10 –1 m = 20 cm 10. Magnetic moment = 0.72 2 Am 2 = M B = 3 0 d M 4 ? ? B H = 18 ?T ? 3 7 d 4 2 72 . 0 10 4 ? ? ? ? ? ? = 18 × 10 –6 ? d 3 = 6 7 10 18 10 414 . 1 72 . 0 ? ? ? ? ? = 0.005656 ? d ˜ 0.2 m = 20 cm 11. The geomagnetic pole is at the end on position of the earth. B = 3 0 d M 2 4 ? ? = 3 3 22 7 ) 10 6400 ( 10 8 2 10 ? ? ? ? ? ˜ 60 × 10 –6 T = 60 ?T 12. B ? = 3.4 × 10 –5 T Given 3 0 R M 4 ? ? = 3.4 × 10 –5 ? M = 7 3 5 10 4 4 R 10 4 . 3 ? ? ? ? ? ? ? ? = 3.4 × 10 2 R 3 B ? at Poles = 3 0 R M 2 4 ? ? = = 6.8 × 10 –5 T 13. ?(dip) = 60° B H = B cos 60° ? B = 52 × 10 –6 = 52 ?T B V = B sin ? = 52 × 10 –6 2 3 = 44.98 ?T ˜ 45 ?T ? 14. If ? 1 and ? 2 be the apparent dips shown by the dip circle in the 2 ?r positions, the true dip ? is given by Cot 2 ? = Cot 2 ? 1 + Cot 2 ? 2 ? Cot 2 ? = Cot 2 45° + Cot 2 53° ? Cot 2 ? = 1.56 ? ? = 38.6 ˜ 39° ? N S P d B M S N B H N W N S S E tan –1 2 N 2 N 1 B H d Permanent Magnets 36.3 15. We know B H = r 2 in 0 ? Give : B H = 3.6 × 10 –5 T ? = 45° ? i = 10 mA = 10 –2 A tan ? = 1 n = ? r = 10 cm = 0.1 m n = i r 2 tan B 0 H ? ? ? = 2 7 1 5 10 10 4 10 1 2 10 6 . 3 ? ? ? ? ? ? ? ? ? ? ? = 0.5732 × 10 3 ˜ 573 turns ? 16. n = 50 A = 2 cm × 2 cm = 2 × 2 × 10 –4 m 2 i = 20 × 10 –3 A B = 0.5 T ? = ? ? B A ni ? ? ? = niAB Sin 90° = 50 × 20 × 10 –3 × 4 × 10 –4 × 0.5 = 2 × 10 –4 NM 17. Given ? = 37° d = 10 cm = 0.1 m We know H B M = ? ? ? ? tan d 2 ) d ( 4 2 2 2 0 ? = ? ? ? ? tan d 2 d 4 4 0 [As the magnet is short] = ? ? ? ? ? ? ? 37 tan 2 ) 1 . 0 ( 10 4 4 3 7 = 0.5 × 0.75 × 1 × 10 –3 × 10 7 = 0.375 × 10 4 = 3.75 ×10 3 Am 2 T –1 18. H B M (found in the previous problem) = 3.75 ×10 3 Am 2 T –1 ? = 37°, d = ? H B M = ? ? ? ? tan ) d ( 4 2 / 3 2 2 0 ? l << d neglecting l w.r.t.d ?? H B M = ? ? ? Tan d 4 3 0 ? 3.75 × 10 3 = 7 10 1 ? × d 3 × 0.75 ? d 3 = 75 . 0 10 10 75 . 3 7 3 ? ? ? = 5 × 10 –4 ? d = 0.079 m = 7.9 cm ? 19. Given H B M = 40 Am 2 /T Since the magnet is short ‘l’ can be neglected So, H B M = 2 d 4 3 0 ? ? ? = 40 ? d 3 = ? ? ? ? ? ? 4 2 10 4 40 7 = 8 × 10 –6 ? d = 2 × 10 –2 m = 2 cm with the northpole pointing towards south. 20. According to oscillation magnetometer, T = H MB 2 ? ? ? 10 ? = 6 4 10 30 M 10 2 . 1 2 ? ? ? ? ? ? ? 2 20 1 ? ? ? ? ? ? = 6 4 10 30 M 10 2 . 1 ? ? ? ? ? ? M = 6 4 10 30 400 10 2 . 1 ? ? ? ? ? = 16 × 10 2 Am 2 = 1600 Am 2 N S Page 4 36.1 CHAPTER – 36 PERMANENT MAGNETS 1. m = 10 Am, d = 5 cm = 0.05 m B = 2 0 r m 4 ? ? = ? ? 2 2 7 10 5 10 10 ? ? ? ? = 25 10 2 ? = 4 × 10 –4 Tesla 2. m 1 =m 2 = 10 Am r = 2 cm = 0.02 m we know Force exerted by tow magnetic poles on each other = 2 2 1 0 r m m 4 ? ? = 4 2 7 10 4 4 10 10 4 ? ? ? ? ? ? ? ? = 2.5 × 10 –2 N 3. B = – ? d dv ? dv = –B dl = – 0.2 × 10 –3 × 0.5 = – 0.1 × 10 –3 Tm Since the sigh is –ve therefore potential decreases. 4. Here dx = 10 sin 30° cm = 5 cm dx dV = B = m 10 5 m T 10 1 . 0 2 4 ? ? ? ? ? Since B is perpendicular to equipotential surface. Here it is at angle 120° with (+ve) xaxis and B = 2 × 10 –4 T 5. B = 2 × 10 –4 T d = 10 cm = 0.1 m (a) if the point at endon postion. B = 3 0 d M 2 4 ? ? ? 2 × 10 –4 = 3 1 7 ) 10 ( M 2 10 ? ? ? ? 2 10 10 10 2 7 3 4 ? ? ? ? ? ? = M ? M = 1 Am 2 (b) If the point is at broadon position 3 0 d M 4 ? ? ? 2 × 10 –4 = 3 1 7 ) 10 ( M 10 ? ? ? ? M = 2 Am 2 6. Given : ? = tan –1 2 ? tan ? = 2 ? 2 = tan 2 ? ? tan ? = 2 cot ? ? 2 tan ? = cot ? We know 2 tan ? = tan ? Comparing we get, tan ? = cot ? or, tan ? = tan(90 – ?) or ? = 90 – ? or ? + ? = 90 Hence magnetic field due to the dipole is ?r to the magnetic axis. ? 7. Magnetic field at the broad side on position : B = ? ? 2 / 3 2 2 0 d M 4 ? ? ? ? 2l = 8 cm d = 3 cm ? 4 × 10 –6 = ? ? 2 / 3 4 4 2 7 10 16 10 9 10 8 m 10 ? ? ? ? ? ? ? ? ? ? ? 4 × 10 –6 = ? ? ? ? 2 / 3 2 / 3 4 9 25 10 8 m 10 ? ? ? ? ? ? m = 9 8 6 10 8 10 125 10 4 ? ? ? ? ? ? ? = 62.5 × 10 –5 Am N S In cm 30° X V 0.1×10 –4 Tm 30° 0.2×10 –4 Tm 0.4×10 –4 Tm 0.3×10 –4 Tm ? ? ? ? P S N ? ? Permanent Magnets 36.2 8. We know for a magnetic dipole with its north pointing the north, the neutral point in the broadside on position. Again B ? in this case = 3 0 d 4 M ? ? ? 3 0 d 4 M ? ? = H B due to earth ? 3 7 d 44 . 1 10 ? ? = 18 ?T ? 3 7 d 44 . 1 10 ? ? = 18 × 10 –6 ? d 3 = 8 × 10 –3 ? d = 2 × 10 –1 m = 20 cm In the plane bisecting the dipole. 9. When the magnet is such that its North faces the geographic south of earth. The neutral point lies along the axial line of the magnet. 3 0 d M 2 4 ? ? = 18 × 10 –6 ? 3 7 d 72 . 0 2 10 ? ? ? = 18 × 10 –6 ? d 3 = 10 18 10 7 . 0 2 6  7 ? ? ? ? ? d = 3 / 1 6 9 10 10 8 ? ? ? ? ? ? ? ? ? ? ? = 2 × 10 –1 m = 20 cm 10. Magnetic moment = 0.72 2 Am 2 = M B = 3 0 d M 4 ? ? B H = 18 ?T ? 3 7 d 4 2 72 . 0 10 4 ? ? ? ? ? ? = 18 × 10 –6 ? d 3 = 6 7 10 18 10 414 . 1 72 . 0 ? ? ? ? ? = 0.005656 ? d ˜ 0.2 m = 20 cm 11. The geomagnetic pole is at the end on position of the earth. B = 3 0 d M 2 4 ? ? = 3 3 22 7 ) 10 6400 ( 10 8 2 10 ? ? ? ? ? ˜ 60 × 10 –6 T = 60 ?T 12. B ? = 3.4 × 10 –5 T Given 3 0 R M 4 ? ? = 3.4 × 10 –5 ? M = 7 3 5 10 4 4 R 10 4 . 3 ? ? ? ? ? ? ? ? = 3.4 × 10 2 R 3 B ? at Poles = 3 0 R M 2 4 ? ? = = 6.8 × 10 –5 T 13. ?(dip) = 60° B H = B cos 60° ? B = 52 × 10 –6 = 52 ?T B V = B sin ? = 52 × 10 –6 2 3 = 44.98 ?T ˜ 45 ?T ? 14. If ? 1 and ? 2 be the apparent dips shown by the dip circle in the 2 ?r positions, the true dip ? is given by Cot 2 ? = Cot 2 ? 1 + Cot 2 ? 2 ? Cot 2 ? = Cot 2 45° + Cot 2 53° ? Cot 2 ? = 1.56 ? ? = 38.6 ˜ 39° ? N S P d B M S N B H N W N S S E tan –1 2 N 2 N 1 B H d Permanent Magnets 36.3 15. We know B H = r 2 in 0 ? Give : B H = 3.6 × 10 –5 T ? = 45° ? i = 10 mA = 10 –2 A tan ? = 1 n = ? r = 10 cm = 0.1 m n = i r 2 tan B 0 H ? ? ? = 2 7 1 5 10 10 4 10 1 2 10 6 . 3 ? ? ? ? ? ? ? ? ? ? ? = 0.5732 × 10 3 ˜ 573 turns ? 16. n = 50 A = 2 cm × 2 cm = 2 × 2 × 10 –4 m 2 i = 20 × 10 –3 A B = 0.5 T ? = ? ? B A ni ? ? ? = niAB Sin 90° = 50 × 20 × 10 –3 × 4 × 10 –4 × 0.5 = 2 × 10 –4 NM 17. Given ? = 37° d = 10 cm = 0.1 m We know H B M = ? ? ? ? tan d 2 ) d ( 4 2 2 2 0 ? = ? ? ? ? tan d 2 d 4 4 0 [As the magnet is short] = ? ? ? ? ? ? ? 37 tan 2 ) 1 . 0 ( 10 4 4 3 7 = 0.5 × 0.75 × 1 × 10 –3 × 10 7 = 0.375 × 10 4 = 3.75 ×10 3 Am 2 T –1 18. H B M (found in the previous problem) = 3.75 ×10 3 Am 2 T –1 ? = 37°, d = ? H B M = ? ? ? ? tan ) d ( 4 2 / 3 2 2 0 ? l << d neglecting l w.r.t.d ?? H B M = ? ? ? Tan d 4 3 0 ? 3.75 × 10 3 = 7 10 1 ? × d 3 × 0.75 ? d 3 = 75 . 0 10 10 75 . 3 7 3 ? ? ? = 5 × 10 –4 ? d = 0.079 m = 7.9 cm ? 19. Given H B M = 40 Am 2 /T Since the magnet is short ‘l’ can be neglected So, H B M = 2 d 4 3 0 ? ? ? = 40 ? d 3 = ? ? ? ? ? ? 4 2 10 4 40 7 = 8 × 10 –6 ? d = 2 × 10 –2 m = 2 cm with the northpole pointing towards south. 20. According to oscillation magnetometer, T = H MB 2 ? ? ? 10 ? = 6 4 10 30 M 10 2 . 1 2 ? ? ? ? ? ? ? 2 20 1 ? ? ? ? ? ? = 6 4 10 30 M 10 2 . 1 ? ? ? ? ? ? M = 6 4 10 30 400 10 2 . 1 ? ? ? ? ? = 16 × 10 2 Am 2 = 1600 Am 2 N S Permanent Magnets 36.4 21. We know : ? = ? ? H mB 2 1 For like poles tied together M = M 1 – M 2 For unlike poles M ? = M 1 + M 2 2 1 ? ? = 2 1 2 1 M M M M ? ? ? 2 2 10 ? ? ? ? ? ? = 2 1 2 1 M M M M ? ? ? 25 = 2 1 2 1 M M M M ? ? ? 24 26 = 2 1 M 2 M 2 ? 2 1 M M = 12 13 22. B H = 24 × 10 –6 T T 1 = 0.1 ? B = B H – B wire = 2.4 × 10 –6 – r i 2 o ? ? = 24 × 10 –6 – 2 . 0 18 10 2 7 ? ? ? = (24 –10) × 10 –6 = 14 × 10 –6 T = H MB 2 ? ? 2 1 T T = H B B ? 2 T 1 . 0 = 6 6 10 24 10 14 ? ? ? ? ? 2 2 T 1 . 0 ? ? ? ? ? ? ? ? = 24 14 ? T 2 2 = 24 14 01 . 0 ? ?T 2 = 0.076 23. T = H MB 2 ? ? Here ?? = 2 ? T 1 = 40 1 min T 2 = ? 2 1 T T = ? ? ? ? 2 T 40 1 = 2 1 ? 2 2 T 1600 1 = 2 1 ? T 2 2 = 800 1 ? T 2 = 0.03536 min For 1 oscillation Time taken = 0.03536 min. For 40 Oscillation Time = 4 × 0.03536 = 1.414 = 2 min 24. ? 1 = 40 oscillations/minute B H = 25 ?T m of second magnet = 1.6 Am 2 d = 20 cm = 0.2 m (a) For north facing north ? 1 = ? ? H MB 2 1 ? 2 = ? ? ? ? ? B B M 2 1 H B = 3 0 d m 4 ? ? ??? 3 7 10 8 6 . 1 10 ? ? ? ? ?= 20 ?T ? 2 1 ? ? = B B B H ? ? 2 40 ? = 5 25 ? ? 2 = 5 40 = 17.88 ˜ 18 osci/min ? (b) For north pole facing south ? ? ???? ? ? H MB 2 1 ? 2 = ? ? ? ? ? B B M 2 1 H 2 1 ? ? = B B B H ? ? 2 40 ? = 45 25 ? ? 2 = ? ? ? ? ? ? 45 25 40 = 53.66 ˜ 54 osci/min ? ? ? ? ?? S N S N N S S NRead More
1. What are permanent magnets? 
2. How are permanent magnets different from temporary magnets? 
3. What are some common uses of permanent magnets? 
4. How can you demagnetize a permanent magnet? 
5. What are some materials that can be used to make permanent magnets? 
130 videos483 docs210 tests

130 videos483 docs210 tests
