RI 16 Electromagnetism Tutorial solutions

RAFFLES INSTITUTION Y5-6 PHYSICS DEPARTMENT 1 Tutorial 16 Electromagnetism Suggested Solutions D1 (a) The magnetic flux density of a magnetic field is numerically equal to the force per unit length per unit current of a long straight conductor at right angles to a uniform magnetic field. ...... (1)FB L= I Taking moments about the axis PQ, sum of anti-clockwise moments = sum of clockwise moments ..... (2) B B NF L Mgx MgxF NL = = Substituting (2) into (1), ( ) 2 Mgx NL MgxB L LN= = Ι I (shown) (b) Rearranging the above equation, 2B LNx Mg= I where emf of battery and resistance of coilE ERRΙ = = = where resistivity of wire total length of wire used to make the coil with turns cross-sectional area of wire R A N A ρ ρ = = = =   (i) When number of turns is doubled , the total length of wire required to form the coil is also doubled. Hence, for the same emf, resistance of the coil is doubled and current flowing through it is halved. For 2N, 2 2 1 22' B LN B LNxx Mg Mg  = = = I I Thus, there is no change in the value of x. (ii) If the length of each side of the coil is halved, the total length of wire required to form the coil would also be halved. Hence, for the same emf, resistance of the coil is halved and current flowing through it is doubled. For 2 L , ( ) 2 22 112' 22 LBN B LNxx Mg Mg   = = =  I I Thus, the value of x is half the original value. P I Q B Mg FB x FB

RAFFLES INSTITUTION Y5-6 PHYSICS DEPARTMENT 2 D2 (a) (i) 1. ( ) ( ) ( )( ) 2 87 22 2 0 10 4 8 10 9 6 10 5 0 10 2 0 10 .LR. .A .. ρ − −− −− × = = × = × ×× Ω 2. V = IR = (50) (9.6 × 10−7) = 4.8 × 10−5 V (ii) 1. By Fleming’s Left Hand Rule, t he force is directed towards the left, perpendicular to the direction of the current and the direction of the magnetic field. 2. F = BIL = (0.12) (50) (2.0 × 10−2) = 0.12 N (iii) ( ) ( ) ( ) 222 0.12 0.032 m 3.2 cm 2.0 10 9.6 10 9.81 Fp hgA Fh Ag ∆ ∆ρ ∆ ρ − = = = = = = ×× (b) Advantage • No moving parts are involved which leads to less wear and tear and also less noise. Disadvantage • The technique can only work with liquids that are conductive. • The circuit needs to handle a large current which may lead to high power losses in the wiring. D3 (a) By the principle of conservation of energy, increase in KE = decrease in electric PE −= = 21 02 2 (shown) Mv QV QVv M (b) The magnetic force on the ions provides for the centripetal force. (c) For an ion with twice the specific charge, the radius of the circular path of the ion in the magnetic field would be reduced. The new radius is given by 2 2 12 MvBQv r Mv M QV MVr BQ BQ M B Q = = = = 12 12MV Vr QBQ B M = = ( ) 12 1' 0. 71 22 Vr r rQB M = = =

RAFFLES INSTITUTION Y5-6 PHYSICS DEPARTMENT 3 D4 (a) (i) Consider the circular motion due to the vertical component of velocity. The magnetic force provides for the centripetal force. ( )( ) ⊥ ⊥ − ⊥ −− = ××° = = =××× 2 31 6 5 19 9.11 10 6.7 10 sin40 0.817 m3.0 10 1.60 10 e e mv BevR mvR Be (ii) ( ) ( ) ππ −− ⊥ = = = ×× ×° 66 6 2 0.8172 1.192 10 s = 1.19 10 s 6.7 10 sin40 RT v (iii) The pitch p depends on the horizontal component of velocity. ( ) ( ) −= = × °×= 66 // 6.7 10 cos 40 1.19 10 6.11 mp vT (or 6.12 m if use 1.192×10−6 s) (b) If the angle θ were very small, v// → v as cos θ → 1 ππ ⊥ ⊥ − = = = = ∴≈ =×× × = 66 Now, from (a)(i), . 22 From (a)(ii), constant 6.7 10 1.19 10 s 7.97 m e e mR v Be mRT v Be p vT p (or 7.99 m if used 1.192×10−6 s) D5 (a) (i) 4.5 1.8 2.5 VR = = = ΩI (ii) ( ) ( ) 23 8 0.60 101.8 2 31.8 m 1.6 10 LR A RAL ρ π ρ − − =   ×  = = = × 2 Total length of wireNumber of turns Circumference 31.8 8.8 10 361 turns or 360 turns (2sf) − = = × = (iii) 2 360Number of turns per metre 3000 (shown)12 10 −= = × (b) ( ) ( )( )Bn I Tµπ −−= = ×= ×73 0 4 10 3000 2.5 9.4 10 (c) (i) ( ) 7 71 along axis 4.0 10 cos30 3.5 10 m sv −= × °= ×

RAFFLES INSTITUTION Y5-6 PHYSICS DEPARTMENT 4 (ii) ( ) 7 71 normal to axis 4.0 10 sin30 2.0 10 m sv −= × °= × (d) The magnetic force on the electron provides for the centripetal force. *Note: The above statement is required to explain your working. 2mvBqv r mvr Bq = = (e) The electron travels in a helical path inside the solenoid. In order for it not to collide, the diameter of the helix must be less than the radius of the solenoid. ( )( ) ( )( ) normal to axis 31 7 3 19 Diameter of helix 2 2 9.11 10 2.0 10 2 9.4 10 1.6 10 0.024 m > 0.014 m (radius of solenoid) r mv Bq − −− = = ×× = ×× = Hence, the electron will collide with the wall of the solenoid. D6 (a) (i) (ii) ( ) ( ) 215.2 10 7.5 0.39 N mF BL −−= = ×=I diameter of helix, 2r radius of solenoid, 0.014 m Viewing into the solenoid (solid circle represents the solenoid, dotted circle represent the helix) v sinθ B field into the page helical path Viewing from the side F

RAFFLES INSTITUTION Y5-6 PHYSICS DEPARTMENT 5 (b) (i) ( )( ) 28 6 23 1 No. of electrons per unit length of wire 7.8 10 1.5 10 1.17 10 m − − = ×× = × 23 24 Force per unit lengthForce on each electron No. of electrons per unit length 0.39 1.17 10 3.3 10 N (shown)− = = × = × (ii) Force on each electron Bqv= ( ) ( )( ) 24 41 2 19 3.3 10 4.0 10 m s 5.2 10 1.6 10 v − −− −− × = = × ×× Alternative method: ( )( )( ) 41 28 6 19 where number density 7.5 4.0 10 m s 7.8 10 1.5 10 1.60 10 nAvq n v −− −− = = = = × ×× × I D7 (a) (i) (ii) increase in KE = decrease in EPE ( ) ( )( ) ( ) 2 19 51 26 1 02 2 2 1.60 10 1400 1.16 10 m s 3.32 10 mv Q V QVv m v − − − −= ∆ ∆= × = = × × + - x x x x x x x x x x x x x x x x x x x x E field B field into the page - + E field B field out of the page

RAFFLES INSTITUTION Y5-6 PHYSICS DEPARTMENT 6 (iii) For the ions to be undeflected, 3 5 6.2 10 0.0534T1.16 10 BEFF Bqv Eq Ev B B = = = ×= = × (b) (i) Charge on neon ions are now +2e, since each atom loses two electrons. From (a) (ii), ( )222 2 1 41eVqVv v' v . vmm ∆∆= = = = The speed of the ions would be greater by a factor of 1.41. (ii) Since the speed of the ions increases, they would experience a larger force due to the magnetic field. The ions would no longer pass through undeflected but are deflected upwards in the region of the two fields. * The curved path is neither parabolic nor circular. D8 (a) increase in KE = decrease in electric PE ( ) ( )( ) ( ) 2 19 51 25 1 02 2 1.60 10 10000 1.25 10 m s 2.04 10 mv Q V v − − − −= ∆ × = = × × (b) (i) ( ) ( ) ( ) 19 15 2 4001.60 10 3. 2 10 N 2.0 10 VF qE q d −− −= = = ×= × × (ii) magnetic force = electric force 1 1 BEFF B qv Eq Ev B = = = where v is the speed of the ions exiting the velocity selector E is the electric field strength within the velocity selector B1 is the magnetic flux density within the velocity selector (perpendicular to E) (iii) ( ) ( ) 2 1 3 400 2.0 10 0.113 T 177 10 EB v −×= = = × + - x x x x x x x x x x x x x x x x x x x x B field into the plane of the paper

RAFFLES INSTITUTION Y5-6 PHYSICS DEPARTMENT 7 (c) (i) The magnetic force acting on an ion is always directed perpendicular to its instantaneous velocity, and this force is constant in magnitude and always pointing towards a fixed point. This leads to a uniform circular motion, where the direction of velocity changes but not its magnitude. (ii) The magnetic force thus provides the centripetal force for the ion to move in a