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