Page 1 Introductory Exercise 23.1 1. [ ] [ ] F F e m = [ ] [ ] qE qvB = Þ E B v é ë ê ù û ú = = - [ ] [L T ] 1 2. F v B ® = ® ´ ® q( ) \ F v ® ^ ® and F B ® ^ ® Because cross product of any two vectors is always perpendicular to both the vectors. 3. No. As F m q ® = ® ´ ® ( ) v B Þ | | sin F m qvB ® = q If F m = 0, either B = 0 or sin q = 0, i.e., q = 0 4. F v B ® = ® ´ ® q( ) = - ´ ´ ´ - - - 4 10 10 10 6 6 2 [( ) ^ ^ ^ 2 3 i j k - + ´ + - ( )] ^ ^ ^ 2 5 3 i j k = - ´ + + - 4 10 4 8 16 2 ( ) ^ ^ ^ i j k = - + + ´ - 16 2 4 10 2 ( ) ^ ^ ^ i j k N Introductory Exercise 23.2 1. As mag netic field can ex ert force on charged par ti cle, it can be ac cel er ated in mag netic field but its speed can not in creases as mag ne tic forc e is al way s per p en dic u lar to the di rec tion of mo tion of charged particle. 2. F v B m e ® = - ® ´ ® ( ) By Flemingâ€™s left hand rule, B ® must be along positive z-axis. 3. As mag netic force pro vides nec es sary cen trip e tal force to the particle to de scribe a circle. qvB mv r = 2 Þ r mv qB = (a) r mv qB = Þ r m µ Hence, electron will describe smaller circle. (b) T r v m qB = = 2 2 p p f T qB m = = 1 2p Þ f m µ 1 \ electron have greater frequency. 4. Elec trons are re fo cused on x-axis at a dis tance equal to pitch, i.e., n p v T = = | | = 2p q mv eB cos Magnetics 23 Page 2 Introductory Exercise 23.1 1. [ ] [ ] F F e m = [ ] [ ] qE qvB = Þ E B v é ë ê ù û ú = = - [ ] [L T ] 1 2. F v B ® = ® ´ ® q( ) \ F v ® ^ ® and F B ® ^ ® Because cross product of any two vectors is always perpendicular to both the vectors. 3. No. As F m q ® = ® ´ ® ( ) v B Þ | | sin F m qvB ® = q If F m = 0, either B = 0 or sin q = 0, i.e., q = 0 4. F v B ® = ® ´ ® q( ) = - ´ ´ ´ - - - 4 10 10 10 6 6 2 [( ) ^ ^ ^ 2 3 i j k - + ´ + - ( )] ^ ^ ^ 2 5 3 i j k = - ´ + + - 4 10 4 8 16 2 ( ) ^ ^ ^ i j k = - + + ´ - 16 2 4 10 2 ( ) ^ ^ ^ i j k N Introductory Exercise 23.2 1. As mag netic field can ex ert force on charged par ti cle, it can be ac cel er ated in mag netic field but its speed can not in creases as mag ne tic forc e is al way s per p en dic u lar to the di rec tion of mo tion of charged particle. 2. F v B m e ® = - ® ´ ® ( ) By Flemingâ€™s left hand rule, B ® must be along positive z-axis. 3. As mag netic force pro vides nec es sary cen trip e tal force to the particle to de scribe a circle. qvB mv r = 2 Þ r mv qB = (a) r mv qB = Þ r m µ Hence, electron will describe smaller circle. (b) T r v m qB = = 2 2 p p f T qB m = = 1 2p Þ f m µ 1 \ electron have greater frequency. 4. Elec trons are re fo cused on x-axis at a dis tance equal to pitch, i.e., n p v T = = | | = 2p q mv eB cos Magnetics 23 5. (a) If L r mv qB ³ = , (b) The particle will describe a semi-circle. Hence, q p = (c) L l =cos q 2 Þ L R 2 2 2 sin cos q q = L R =sinq Þ sin q = 1 2 Þ q p = 6 6. r mv eB mk eB = = 2 = = 2 1 2 meV eB B mV e For electron, r = ´ ´ ´ ´ - - 1 2 10 100 10 31 19 0.2 9.1 1.6 = ´ - 1.67 10 4 m = 0.0167 cm For proton r = ´ ´ ´ ´ - - 1 2 10 100 10 27 19 0.2 1.67 1.6 = ´ - 7 10 3 m = 0.7 cm 7. r mv qB m k qB = = 2 \ r m q µ \ r r r p d : : : : a = 1 1 2 1 4 2 = 1 2 1 : : Introductory Exercise 23.3 1. Let at any in stant V i j k ® = + + V V V x y z ^ ^ ^ Now, V V V x y 2 2 0 2 + = = constant and V V qE m f 2 0 = - V ® is minimum when V 2 0 = at f mv qE = 0 and V V min = 0 2. Af ter one rev o lu tion, y = 0 , x p = = pitch of heating = 2p q mv qB sin Hence, coordination of the particle, = ( , ) x b = æ è ç ç ö ø ÷ ÷ 0 2 , sin p q mv qB 3. F l B i j k ® = ® ´ ® = ´ + i ilB ( ) [ ( )] ^ ^ ^ ( ) F ® = 2 ilB 4. No. as i i j k i j i j k ^ ^ ^ ^ ^ ^ ^ ^ ^ ( ) ( ) ´ + + = ´ + ´ ´ But i j ^ ^ ´ = 0 \ i i j k i j k ^ ^ ^ ^ ^ ^ ^ ( ) ( ) ´ + + = ´ ´ 84 × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × O q q × × × q/2 p 2 â€“ q 2 L Page 3 Introductory Exercise 23.1 1. [ ] [ ] F F e m = [ ] [ ] qE qvB = Þ E B v é ë ê ù û ú = = - [ ] [L T ] 1 2. F v B ® = ® ´ ® q( ) \ F v ® ^ ® and F B ® ^ ® Because cross product of any two vectors is always perpendicular to both the vectors. 3. No. As F m q ® = ® ´ ® ( ) v B Þ | | sin F m qvB ® = q If F m = 0, either B = 0 or sin q = 0, i.e., q = 0 4. F v B ® = ® ´ ® q( ) = - ´ ´ ´ - - - 4 10 10 10 6 6 2 [( ) ^ ^ ^ 2 3 i j k - + ´ + - ( )] ^ ^ ^ 2 5 3 i j k = - ´ + + - 4 10 4 8 16 2 ( ) ^ ^ ^ i j k = - + + ´ - 16 2 4 10 2 ( ) ^ ^ ^ i j k N Introductory Exercise 23.2 1. As mag netic field can ex ert force on charged par ti cle, it can be ac cel er ated in mag netic field but its speed can not in creases as mag ne tic forc e is al way s per p en dic u lar to the di rec tion of mo tion of charged particle. 2. F v B m e ® = - ® ´ ® ( ) By Flemingâ€™s left hand rule, B ® must be along positive z-axis. 3. As mag netic force pro vides nec es sary cen trip e tal force to the particle to de scribe a circle. qvB mv r = 2 Þ r mv qB = (a) r mv qB = Þ r m µ Hence, electron will describe smaller circle. (b) T r v m qB = = 2 2 p p f T qB m = = 1 2p Þ f m µ 1 \ electron have greater frequency. 4. Elec trons are re fo cused on x-axis at a dis tance equal to pitch, i.e., n p v T = = | | = 2p q mv eB cos Magnetics 23 5. (a) If L r mv qB ³ = , (b) The particle will describe a semi-circle. Hence, q p = (c) L l =cos q 2 Þ L R 2 2 2 sin cos q q = L R =sinq Þ sin q = 1 2 Þ q p = 6 6. r mv eB mk eB = = 2 = = 2 1 2 meV eB B mV e For electron, r = ´ ´ ´ ´ - - 1 2 10 100 10 31 19 0.2 9.1 1.6 = ´ - 1.67 10 4 m = 0.0167 cm For proton r = ´ ´ ´ ´ - - 1 2 10 100 10 27 19 0.2 1.67 1.6 = ´ - 7 10 3 m = 0.7 cm 7. r mv qB m k qB = = 2 \ r m q µ \ r r r p d : : : : a = 1 1 2 1 4 2 = 1 2 1 : : Introductory Exercise 23.3 1. Let at any in stant V i j k ® = + + V V V x y z ^ ^ ^ Now, V V V x y 2 2 0 2 + = = constant and V V qE m f 2 0 = - V ® is minimum when V 2 0 = at f mv qE = 0 and V V min = 0 2. Af ter one rev o lu tion, y = 0 , x p = = pitch of heating = 2p q mv qB sin Hence, coordination of the particle, = ( , ) x b = æ è ç ç ö ø ÷ ÷ 0 2 , sin p q mv qB 3. F l B i j k ® = ® ´ ® = ´ + i ilB ( ) [ ( )] ^ ^ ^ ( ) F ® = 2 ilB 4. No. as i i j k i j i j k ^ ^ ^ ^ ^ ^ ^ ^ ^ ( ) ( ) ´ + + = ´ + ´ ´ But i j ^ ^ ´ = 0 \ i i j k i j k ^ ^ ^ ^ ^ ^ ^ ( ) ( ) ´ + + = ´ ´ 84 × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × O q q × × × q/2 p 2 â€“ q 2 L Introductory Exercise 23.4 1. Con sider the disc to be made up of large num ber of elementary concentric rings. Con sider one such ring of ra dius x and thick ne ss d x . Charge on this ring dq q R dA q R x dx = × + ´ p p p 2 2 2 dq qxdx R = 2 2 Current in this ring, di dq T dq qx dx R = + = w p w p 2 2 \ Magnetic moment of this ring, dM di A qx dx R x = ´ = ´ w p p 2 2 = wq R x dx 2 3 \ Magnetic moment of entire disc, M dM q R x dx R = = ò ò w 2 3 0 = é ë ê ê ù û ú ú = w w q R R qR 2 4 2 4 1 4 2. M OA AB ® = ´ ® ´ ® i [( )] OA j k ® = + OA OA cos sin ^ ^ q q AB i ® = AB ^ \ M j k i ® = ´ × + ´ i OA AB [(cos sin ) ] ^ ^ ^ q q = ´ ´ + æ è ç ç ö ø ÷ ÷ ´ é ë ê ê ù û ú ú 4 3 2 1 2 0.2 0.1 j k i ^ ^ ^ = - ( ) ^ ^ 0.04 0.07 j k A-m 2 Introductory Exercise 23.5 1. (a) B B B B 1 2 3 4 = = = = × ° + ° m p 0 4 2 45 45 i l / [sin sin ] = × m p 0 4 2 2i l Net magnetic field at the centre of the square, B B B B B i l = + + + = × 1 2 3 4 0 4 8 2 m p = = 2 2 0 m p m i l 28.3 T (inward) (b)If the conductor is converted into a circular loop, then 2 4 pr l = Þ r l = 2 p B i r i l = = = m p m m 0 0 2 4 24.7 T (inward) 2. B i x = × m p 0 4 (As P is lying near one end of conductor 1) B 2 0 = (Magnetic field on the axis of a current carrying conductor is zero) B B = 1 = × m p 0 4 i x By right hand thumb rule, direction of magnetic field at P is inward. 85 1 2 45° i 3 l l 2 4 2 i x P i 1 x Page 4 Introductory Exercise 23.1 1. [ ] [ ] F F e m = [ ] [ ] qE qvB = Þ E B v é ë ê ù û ú = = - [ ] [L T ] 1 2. F v B ® = ® ´ ® q( ) \ F v ® ^ ® and F B ® ^ ® Because cross product of any two vectors is always perpendicular to both the vectors. 3. No. As F m q ® = ® ´ ® ( ) v B Þ | | sin F m qvB ® = q If F m = 0, either B = 0 or sin q = 0, i.e., q = 0 4. F v B ® = ® ´ ® q( ) = - ´ ´ ´ - - - 4 10 10 10 6 6 2 [( ) ^ ^ ^ 2 3 i j k - + ´ + - ( )] ^ ^ ^ 2 5 3 i j k = - ´ + + - 4 10 4 8 16 2 ( ) ^ ^ ^ i j k = - + + ´ - 16 2 4 10 2 ( ) ^ ^ ^ i j k N Introductory Exercise 23.2 1. As mag netic field can ex ert force on charged par ti cle, it can be ac cel er ated in mag netic field but its speed can not in creases as mag ne tic forc e is al way s per p en dic u lar to the di rec tion of mo tion of charged particle. 2. F v B m e ® = - ® ´ ® ( ) By Flemingâ€™s left hand rule, B ® must be along positive z-axis. 3. As mag netic force pro vides nec es sary cen trip e tal force to the particle to de scribe a circle. qvB mv r = 2 Þ r mv qB = (a) r mv qB = Þ r m µ Hence, electron will describe smaller circle. (b) T r v m qB = = 2 2 p p f T qB m = = 1 2p Þ f m µ 1 \ electron have greater frequency. 4. Elec trons are re fo cused on x-axis at a dis tance equal to pitch, i.e., n p v T = = | | = 2p q mv eB cos Magnetics 23 5. (a) If L r mv qB ³ = , (b) The particle will describe a semi-circle. Hence, q p = (c) L l =cos q 2 Þ L R 2 2 2 sin cos q q = L R =sinq Þ sin q = 1 2 Þ q p = 6 6. r mv eB mk eB = = 2 = = 2 1 2 meV eB B mV e For electron, r = ´ ´ ´ ´ - - 1 2 10 100 10 31 19 0.2 9.1 1.6 = ´ - 1.67 10 4 m = 0.0167 cm For proton r = ´ ´ ´ ´ - - 1 2 10 100 10 27 19 0.2 1.67 1.6 = ´ - 7 10 3 m = 0.7 cm 7. r mv qB m k qB = = 2 \ r m q µ \ r r r p d : : : : a = 1 1 2 1 4 2 = 1 2 1 : : Introductory Exercise 23.3 1. Let at any in stant V i j k ® = + + V V V x y z ^ ^ ^ Now, V V V x y 2 2 0 2 + = = constant and V V qE m f 2 0 = - V ® is minimum when V 2 0 = at f mv qE = 0 and V V min = 0 2. Af ter one rev o lu tion, y = 0 , x p = = pitch of heating = 2p q mv qB sin Hence, coordination of the particle, = ( , ) x b = æ è ç ç ö ø ÷ ÷ 0 2 , sin p q mv qB 3. F l B i j k ® = ® ´ ® = ´ + i ilB ( ) [ ( )] ^ ^ ^ ( ) F ® = 2 ilB 4. No. as i i j k i j i j k ^ ^ ^ ^ ^ ^ ^ ^ ^ ( ) ( ) ´ + + = ´ + ´ ´ But i j ^ ^ ´ = 0 \ i i j k i j k ^ ^ ^ ^ ^ ^ ^ ( ) ( ) ´ + + = ´ ´ 84 × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × O q q × × × q/2 p 2 â€“ q 2 L Introductory Exercise 23.4 1. Con sider the disc to be made up of large num ber of elementary concentric rings. Con sider one such ring of ra dius x and thick ne ss d x . Charge on this ring dq q R dA q R x dx = × + ´ p p p 2 2 2 dq qxdx R = 2 2 Current in this ring, di dq T dq qx dx R = + = w p w p 2 2 \ Magnetic moment of this ring, dM di A qx dx R x = ´ = ´ w p p 2 2 = wq R x dx 2 3 \ Magnetic moment of entire disc, M dM q R x dx R = = ò ò w 2 3 0 = é ë ê ê ù û ú ú = w w q R R qR 2 4 2 4 1 4 2. M OA AB ® = ´ ® ´ ® i [( )] OA j k ® = + OA OA cos sin ^ ^ q q AB i ® = AB ^ \ M j k i ® = ´ × + ´ i OA AB [(cos sin ) ] ^ ^ ^ q q = ´ ´ + æ è ç ç ö ø ÷ ÷ ´ é ë ê ê ù û ú ú 4 3 2 1 2 0.2 0.1 j k i ^ ^ ^ = - ( ) ^ ^ 0.04 0.07 j k A-m 2 Introductory Exercise 23.5 1. (a) B B B B 1 2 3 4 = = = = × ° + ° m p 0 4 2 45 45 i l / [sin sin ] = × m p 0 4 2 2i l Net magnetic field at the centre of the square, B B B B B i l = + + + = × 1 2 3 4 0 4 8 2 m p = = 2 2 0 m p m i l 28.3 T (inward) (b)If the conductor is converted into a circular loop, then 2 4 pr l = Þ r l = 2 p B i r i l = = = m p m m 0 0 2 4 24.7 T (inward) 2. B i x = × m p 0 4 (As P is lying near one end of conductor 1) B 2 0 = (Magnetic field on the axis of a current carrying conductor is zero) B B = 1 = × m p 0 4 i x By right hand thumb rule, direction of magnetic field at P is inward. 85 1 2 45° i 3 l l 2 4 2 i x P i 1 x 3. Mag netic field due to straight con duc tor at O B i R 1 0 4 2 = × m p Magnetic field at O due to circular loop B i R 2 0 2 = m By right hand thumb rule, both the filds are acting inward. Hence, B B B i R = + = + 1 2 0 2 1 m p p ( ) = ´ ´ ´ ´ + æ è ç ö ø ÷ - - 4 10 7 2 10 10 1 22 7 7 2 p p = ´ = - 58 10 58 6 T T m (inward). 4. B B 1 2 0 = = (Mag netic field on the axis of cur rent car ry ing con duc t or is zero) B i R i R 3 0 0 1 4 2 8 = × = m m = ´ ´ ´ ´ - - 4 10 5 8 3 10 7 2 p = ´ - 2.62 10 5 T =26.2 T m (inward). 5. B B 1 2 0 = = (Mag netic field on the axis of straight con duc tor is zero) B i b i b 2 0 0 60 360 2 12 = ° ° × = m m (inward) B i a a 3 0 0 60 360 2 12 = ° ° = = m m (outward) As B B 3 2 > , Net magnetic field at P, B B B = - 3 2 = - é ë ê ù û ú m 0 12 1 1 i a b 6. AB AP , and BP from Pythagorus trip let, hence Ð = ° APB 90 \ B PB 1 0 1 1 4 2 ® = × m p i r ^ B AP 2 0 2 2 4 2 ® = × m p i r ^ B B B = + 1 2 2 2 = æ è ç ç ö ø ÷ ÷ + æ è ç ç ö ø ÷ ÷ m p 0 1 1 2 2 2 2 2 i r i r = ´ ´ æ è ç ö ø ÷ + æ è ç ö ø ÷ - 4 10 2 3 3 7 2 2 p p 0.05 0.12 = ´ - 1.3 10 5 T =13 mT 7. t q = NIABcos = ´ ´ ´ ´ ´ ° 100 3 30 1.2 0.4 0. 0.8 cos = 9.98 N-m Rotation will be clockwise as seen from above. 86 60° 2 3 1 4 X X B A i 1 i 2 P ® B 2 ® B 1 5.0cm 13.0cm 12.0 cm i O R 3 O i 1 2 Page 5 Introductory Exercise 23.1 1. [ ] [ ] F F e m = [ ] [ ] qE qvB = Þ E B v é ë ê ù û ú = = - [ ] [L T ] 1 2. F v B ® = ® ´ ® q( ) \ F v ® ^ ® and F B ® ^ ® Because cross product of any two vectors is always perpendicular to both the vectors. 3. No. As F m q ® = ® ´ ® ( ) v B Þ | | sin F m qvB ® = q If F m = 0, either B = 0 or sin q = 0, i.e., q = 0 4. F v B ® = ® ´ ® q( ) = - ´ ´ ´ - - - 4 10 10 10 6 6 2 [( ) ^ ^ ^ 2 3 i j k - + ´ + - ( )] ^ ^ ^ 2 5 3 i j k = - ´ + + - 4 10 4 8 16 2 ( ) ^ ^ ^ i j k = - + + ´ - 16 2 4 10 2 ( ) ^ ^ ^ i j k N Introductory Exercise 23.2 1. As mag netic field can ex ert force on charged par ti cle, it can be ac cel er ated in mag netic field but its speed can not in creases as mag ne tic forc e is al way s per p en dic u lar to the di rec tion of mo tion of charged particle. 2. F v B m e ® = - ® ´ ® ( ) By Flemingâ€™s left hand rule, B ® must be along positive z-axis. 3. As mag netic force pro vides nec es sary cen trip e tal force to the particle to de scribe a circle. qvB mv r = 2 Þ r mv qB = (a) r mv qB = Þ r m µ Hence, electron will describe smaller circle. (b) T r v m qB = = 2 2 p p f T qB m = = 1 2p Þ f m µ 1 \ electron have greater frequency. 4. Elec trons are re fo cused on x-axis at a dis tance equal to pitch, i.e., n p v T = = | | = 2p q mv eB cos Magnetics 23 5. (a) If L r mv qB ³ = , (b) The particle will describe a semi-circle. Hence, q p = (c) L l =cos q 2 Þ L R 2 2 2 sin cos q q = L R =sinq Þ sin q = 1 2 Þ q p = 6 6. r mv eB mk eB = = 2 = = 2 1 2 meV eB B mV e For electron, r = ´ ´ ´ ´ - - 1 2 10 100 10 31 19 0.2 9.1 1.6 = ´ - 1.67 10 4 m = 0.0167 cm For proton r = ´ ´ ´ ´ - - 1 2 10 100 10 27 19 0.2 1.67 1.6 = ´ - 7 10 3 m = 0.7 cm 7. r mv qB m k qB = = 2 \ r m q µ \ r r r p d : : : : a = 1 1 2 1 4 2 = 1 2 1 : : Introductory Exercise 23.3 1. Let at any in stant V i j k ® = + + V V V x y z ^ ^ ^ Now, V V V x y 2 2 0 2 + = = constant and V V qE m f 2 0 = - V ® is minimum when V 2 0 = at f mv qE = 0 and V V min = 0 2. Af ter one rev o lu tion, y = 0 , x p = = pitch of heating = 2p q mv qB sin Hence, coordination of the particle, = ( , ) x b = æ è ç ç ö ø ÷ ÷ 0 2 , sin p q mv qB 3. F l B i j k ® = ® ´ ® = ´ + i ilB ( ) [ ( )] ^ ^ ^ ( ) F ® = 2 ilB 4. No. as i i j k i j i j k ^ ^ ^ ^ ^ ^ ^ ^ ^ ( ) ( ) ´ + + = ´ + ´ ´ But i j ^ ^ ´ = 0 \ i i j k i j k ^ ^ ^ ^ ^ ^ ^ ( ) ( ) ´ + + = ´ ´ 84 × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × O q q × × × q/2 p 2 â€“ q 2 L Introductory Exercise 23.4 1. Con sider the disc to be made up of large num ber of elementary concentric rings. Con sider one such ring of ra dius x and thick ne ss d x . Charge on this ring dq q R dA q R x dx = × + ´ p p p 2 2 2 dq qxdx R = 2 2 Current in this ring, di dq T dq qx dx R = + = w p w p 2 2 \ Magnetic moment of this ring, dM di A qx dx R x = ´ = ´ w p p 2 2 = wq R x dx 2 3 \ Magnetic moment of entire disc, M dM q R x dx R = = ò ò w 2 3 0 = é ë ê ê ù û ú ú = w w q R R qR 2 4 2 4 1 4 2. M OA AB ® = ´ ® ´ ® i [( )] OA j k ® = + OA OA cos sin ^ ^ q q AB i ® = AB ^ \ M j k i ® = ´ × + ´ i OA AB [(cos sin ) ] ^ ^ ^ q q = ´ ´ + æ è ç ç ö ø ÷ ÷ ´ é ë ê ê ù û ú ú 4 3 2 1 2 0.2 0.1 j k i ^ ^ ^ = - ( ) ^ ^ 0.04 0.07 j k A-m 2 Introductory Exercise 23.5 1. (a) B B B B 1 2 3 4 = = = = × ° + ° m p 0 4 2 45 45 i l / [sin sin ] = × m p 0 4 2 2i l Net magnetic field at the centre of the square, B B B B B i l = + + + = × 1 2 3 4 0 4 8 2 m p = = 2 2 0 m p m i l 28.3 T (inward) (b)If the conductor is converted into a circular loop, then 2 4 pr l = Þ r l = 2 p B i r i l = = = m p m m 0 0 2 4 24.7 T (inward) 2. B i x = × m p 0 4 (As P is lying near one end of conductor 1) B 2 0 = (Magnetic field on the axis of a current carrying conductor is zero) B B = 1 = × m p 0 4 i x By right hand thumb rule, direction of magnetic field at P is inward. 85 1 2 45° i 3 l l 2 4 2 i x P i 1 x 3. Mag netic field due to straight con duc tor at O B i R 1 0 4 2 = × m p Magnetic field at O due to circular loop B i R 2 0 2 = m By right hand thumb rule, both the filds are acting inward. Hence, B B B i R = + = + 1 2 0 2 1 m p p ( ) = ´ ´ ´ ´ + æ è ç ö ø ÷ - - 4 10 7 2 10 10 1 22 7 7 2 p p = ´ = - 58 10 58 6 T T m (inward). 4. B B 1 2 0 = = (Mag netic field on the axis of cur rent car ry ing con duc t or is zero) B i R i R 3 0 0 1 4 2 8 = × = m m = ´ ´ ´ ´ - - 4 10 5 8 3 10 7 2 p = ´ - 2.62 10 5 T =26.2 T m (inward). 5. B B 1 2 0 = = (Mag netic field on the axis of straight con duc tor is zero) B i b i b 2 0 0 60 360 2 12 = ° ° × = m m (inward) B i a a 3 0 0 60 360 2 12 = ° ° = = m m (outward) As B B 3 2 > , Net magnetic field at P, B B B = - 3 2 = - é ë ê ù û ú m 0 12 1 1 i a b 6. AB AP , and BP from Pythagorus trip let, hence Ð = ° APB 90 \ B PB 1 0 1 1 4 2 ® = × m p i r ^ B AP 2 0 2 2 4 2 ® = × m p i r ^ B B B = + 1 2 2 2 = æ è ç ç ö ø ÷ ÷ + æ è ç ç ö ø ÷ ÷ m p 0 1 1 2 2 2 2 2 i r i r = ´ ´ æ è ç ö ø ÷ + æ è ç ö ø ÷ - 4 10 2 3 3 7 2 2 p p 0.05 0.12 = ´ - 1.3 10 5 T =13 mT 7. t q = NIABcos = ´ ´ ´ ´ ´ ° 100 3 30 1.2 0.4 0. 0.8 cos = 9.98 N-m Rotation will be clockwise as seen from above. 86 60° 2 3 1 4 X X B A i 1 i 2 P ® B 2 ® B 1 5.0cm 13.0cm 12.0 cm i O R 3 O i 1 2 Introductory Exercise 23.6 1. By right hand thumb rule, di rec tion of magnetic field due to con duc tor A B C , , and D are as shown in fig ure. B B B B I r A B C D = = = = × m p 0 4 2 Here, I = 5 A r a = = = 2 2 0.2 0.14 \ Net magnetic field at P B B B B B A D B C = + + + ( ) ( ) 2 2 = × m p 0 4 4 2 I r = ´ ´ = ´ - - 10 4 2 5 20 10 7 6 0.2 / 2 T = 20 mT Clearly resultant magnetic field is downward. 2. At point A B I r 1 0 1 1 4 = × m p B 2 0 = (Magnetic field inside a current carrying hollow cylinder is zero) \ B B B I r a = + = × 1 2 0 1 1 4 m p = ´ ´ - - - - 10 1 1 10 10 7 3 4 T =100mT (upward) At point B B I r 1 0 0 1 2 4 = × m pe , B I r 2 0 2 2 4 = × ¯ m p Net field at B B B B r I I = - = × - 2 1 0 2 2 1 4 1 m p ( ) = ´ ´ - = ´ - - - 10 3 10 3 3 2 10 7 4 ( ) 0.67 T = 67 mT 3. Con sider the cyl in der to be made up of large num ber of el e men tary hol low cyl in ders. Consider one such cylinder of radius r and thickness dr. Current passing through this hollow cylinder, di jdA j rdr brdr = = ( ) 2 2 2 p p (a) Total current inside the portion of radius r 1 , I di b r dr r 1 2 0 2 1 = = ò ò p = é ë ê ê ù û ú ú 2 3 1 3 0 1 pb r r = 2 3 1 3 pbr By ampereâ€™s circuital law, B dl i r × = ò 2 01 1 p m B r br 1 1 0 1 3 2 2 3 ´ = æ è ç ö ø ÷ p m p Þ B br 1 0 1 2 3 = m (b) Total current inside the cylinder i b r dr R = ò 2 2 0 p = 2 3 3 pbR B i r bR r 2 0 2 0 3 2 4 2 3 = = m p m 87 R r I 2 × × I 1 b B 1 B 2 q m ® B ® B A ® B D ® B B ® B C A X 2m X B D CRead More

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### Chapter 24 - Electromagnetic Induction (Part - 1) - Physics, Solution by D C Pandey, NEET

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### Chapter 24 - Electromagnetic Induction (Part - 2) - Physics, Solution by D C Pandey, NEET

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