DHS 16 Electromagnetism (Tutorial Answers)

Dunman High School (Senior High) H2 Physics Topic 16: Electromagnetism Tutorial Solutions Topic 16: Electromagnetism Page 1 of 8 Tutorial Solutions spacing increases with distance from wire to indicate weaker field Magnetic Fields 1. Long Straight Wire Flat Circular Coil Long Solenoid 2. (a) Currents in the same direction (b) Currents in opposite directions 3. i. Directed to the left (by the right hand grip rule) ii Current carrying conductor should be placed parallel to the field lines. Parallel equidistant lines to indicate uniform field lines spacing wider to indicate weaker field strength lines more closely spaced to indicate stronger field strength lines spacing wider to indicate weaker field strength lines more closely spaced to indicate stronger field strength

Topic 16: Electromagnetism Page 2 of 8 Tutorial Solutions Magnetic force on current-carrying conductors 4. D. Magnetic field is parallel to motion of electron, θ is either 0° or 180°. Using F = BIL sinθ, no electromagnetic force is experienced by the electron. 5. C. Using Fleming’s left hand rule. 6. ---------(1) 7. Ans: B. Wires X and Z each exert a leftward force on Y. 8. Ans: A. Use the Right Hand G rip rule at each wire to find the direction of magnetic field due to each wire on point O. Thereafter, do a vector sum of all 4 magnetic field vectors to confirm that the resultant field points to the left. 9. i Let BF = CE = d and BC = FE = L For rotational equilibrium, Anti-clockwise moment due to magnetic force = Clockwise moment due to weight 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵90𝑜𝑜 × 𝑑𝑑 2 = 𝑚𝑚𝑚𝑚 𝑑𝑑 2 𝐵𝐵 = 𝑚𝑚𝑚𝑚 𝐵𝐵𝐵𝐵 ii 𝐵𝐵 = 𝑚𝑚𝑚𝑚 𝐵𝐵𝐵𝐵 𝜇𝜇𝑜𝑜𝐵𝐵𝐵𝐵 = 𝑚𝑚𝑚𝑚 𝐵𝐵𝐵𝐵 𝐵𝐵 = � 𝑚𝑚𝑚𝑚 𝜇𝜇𝑜𝑜𝐵𝐵𝐵𝐵 = � (0.10 × 10−3)(9.81) (4𝜋𝜋 × 10−7)(1800)(0.025) = 4.17 A oBILF 90sin1 = BIL=×⇒ −3100.8 ( )°    = 60sin2 2 LIBF 2 60sin100.8 3 °××= − mN 5.3N105.3 3 =×= −

Topic 16: Electromagnetism Page 3 of 8 Tutorial Solutions 10. With no current, the reading of 142.0 g is the mass of the horse shoe magnet. With a current of 2.0 A in the wire from X to Y, a magnetic force F B acts upwards on the wire XY. By Newton’s 3rd Law, there is an equal and opposite (downwards) force on the pan, causing the reading on the top pan balance to increase. When a current of 3.0 A flows from Y to X, the magnetic force on the wire will act downwards. By Newton’s 3rd Law, there is an equal and opposite (upwards) force on the pan, causing the reading on the top pan balance to decrease. 11 (a) (i) Force per unit length = BIsinθ (ii) By Fleming’s LHR, the force is directed into the plane of the paper. (b) (i) Magnetic field due to current in the upper loop is normal to the current in the lower loop. This causes a magnetic force to be exerted on the lower loop. Based on Newton’s 3 rd law, a magnetic force is also exerted on the upper loop which is of the same magnitude but opposite in direction. Since currents are in the same direction for both loops , by Fleming’s left hand rule, the forces are attractive and the loops attract each other. (ii) B = 2 x 10-7 I / (0.75 x 10-2) = 2.67 x 10-5 I Force by weight = 0.26 x 10-3 x 9.81 = 2.55 x 10-3 N Force by weight = magnetic force exerted on the loop = BIL 2.55 x 10-3 = (2.67 x 10-5 I)(I)(2π x 4.7 x 10-2) I = 18 A ( ) 81.9100.1426.14490sin 3 ××−=°= −BILFB ( ) ( ) 81.9101426.1440.2 3 ××−= −LB 0127.0=⇒ BL ( ) °= 90sin0.3' LBFB ( )( )( ) ( )0127.0381.9100.142 3' ×=− −m g 1.138' =⇒ m

Topic 16: Electromagnetism Page 4 of 8 Tutorial Solutions Magnetic force on moving charges 12 • Electrons move towards observer’s eye. • Hence, conventional current is into the plane of this paper. • Using Fleming’s LH rule, electron deflects to the left (due to the magnetic field) as shown above. 13 i Assuming that the charged particles enter at an angle 90o to the magnetic field. Since the magnetic force on the charged particle provides the centripetal force, By Newton’s 2nd law: 𝐹𝐹𝐵𝐵 = 𝐹𝐹𝐶𝐶 𝐵𝐵𝐵𝐵𝐵𝐵 sin90o = 𝑚𝑚 𝐵𝐵2 𝑟𝑟 𝐵𝐵 = 𝐵𝐵𝐵𝐵𝑟𝑟 𝑚𝑚 ii If the particles were electrons, 𝐵𝐵 = (6.0 × 10−5)(1.6 × 10−19)(50000) (9.11 × 10−31) = 5.27 × 1011 m s−1 Since no particles travels faster than the speed of light (3.0 × 108 m s-1), it is not possible for the particles to be electrons. iii Considering the proton: 𝐵𝐵 = (6.0 × 10−5)(1.6 × 10−19)(50000) (1.67 × 10−27) = 2.87 × 108 m s−1 < 𝑐𝑐 Hence one possible particle that could have caused the aurora could be the proton. Undeflected Beam Deflection due to B-Field Deflection due to E-Field 45o

Topic 16: Electromagnetism Page 5 of 8 Tutorial Solutions 14 (a) Clockwise circular path passing through P. (b) i 𝐹𝐹 = 𝐵𝐵𝐵𝐵𝐵𝐵sin𝜃𝜃 = (1.5 × 10−3)(1.60 × 10−19)(2.9 × 107)sin90o = 6.96 × 10−15 N ii Since the magnetic force on the electron provides the centripetal force, 𝐹𝐹𝐵𝐵 = 𝐹𝐹𝐶𝐶 𝐵𝐵𝐵𝐵𝐵𝐵 = 𝑚𝑚 𝐵𝐵2 𝑟𝑟 ∴ 𝑟𝑟 = 𝑚𝑚 𝐵𝐵 𝐵𝐵𝐵𝐵 = (9.11 × 10−31)(2.9 × 107) (1.5 × 10−3)(1.60 × 10−19) = 0.11 m iii Time for half period = dist ÷ speed = (π × 0.11) ÷ (2.9 × 107) = 1.19 × 10-8 s iv Kinetic Energy = 1 2 𝑚𝑚𝐵𝐵2 = 1 2 (9.11 × 10−31)(2.9 × 107)2 = 3.83 × 10−16 J (c) Loss in Electric P.E. = Gain in K.E. of the electron e∆V = 3.83 × 10−16 ∆V = 2390 V 15 (a) i = 9.6 × 10-7 Ω ii V = I R = (50)(9.6 × 10-7) = 4.8 × 10-5 V AR  ρ= ) 02 . 0 05 . 0 ( ) 02 . 0 )( 10 8 . 4 (8 × ×= −

Topic 16: Electromagnetism Page 6 of 8 Tutorial Solutions (b) i Sodium positive ions flow in the direction of the conventional current. By Fleming’s Left hand rule, the magnetic force exerted on the liquid is directed to the left. ii F = sinθ = (0.12)(50)(0.02) sin90° = 0.12 N (c) The electromagnetic force acting from right to left exerts a pressure on the cross- sectional area of the horizontal tube. Pressure = This pressure causes the height of the liquid in the left column to be greater than the height of the liquid in the right column by ⇒ ⇒ m (d) Advantage: It involves no moving physical parts and hence little wear and tear; maintenance free. Or less noise Disadvantage: It involves the need to handle relatively high currents which poses a safety hazard. It can only be used if the fluid is highly conductive. BIL 020 . 0 020 . 0 12 . 0 Area Force ×= 020 . 0 020 . 0 12 . 0 ×= × × ∆g hρ 020 . 0 020 . 0 12 . 081 . 9 10 6 . 92 ×= × × × ∆h 032 . 0= ∆h 2.0 cm 50 A magnetic field

Topic 16: Electromagnetism Page 7 of 8 Tutorial Solutions Crossed Fields 16 (a) Loss in Electric P.E. = Gain in K.E. 𝑄𝑄𝑄𝑄 = 1 2 𝑀𝑀𝐵𝐵2 𝐵𝐵 = �2𝑄𝑄𝑄𝑄 𝑀𝑀 (b) i Leftwards. Electric field must be directed to the left so that the electric force on the negative charge is directed to the right (opposite in direction to the electromagnetic force). ii Based on Fleming’s LHR, the electromagnetic force on the negatively charged particles is directed to the left. With the electric field directed to the left, the electric force on the negatively charged particles is directed to the right. When these two forces are of the same magnitude (but in opposite directions), the resultant will be zero. Based on Newton’s 1 st law, t he particles will travel with constant velocity in a straight line. Because the electromagnetic force depends on the speed of the charged particles, there is only one speed in which the electromagnetic force acted on the charged particles is equal to the electric force. iii In order for the particles to pass through undeviated, 𝐹𝐹𝐵𝐵 = 𝐹𝐹𝐸𝐸 𝐵𝐵1𝑄𝑄𝐵𝐵 = 𝑄𝑄𝑄𝑄 𝐵𝐵 = 𝑄𝑄 𝐵𝐵1

Topic 16