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