RI Y6 Remedial EM Soln

Raffles Institution Year 5-6 Physics Department 1 2025 Year 6 H2 Physics Remedial Chapter 16: Electromagnetism Suggested Solutions 1 (a) (i) (ii) Magnetic force F on wire X and wire Y as indicated. (iii) 1 . BX due to Y = [(4π x 10-7)(3.0)] / [2π (0.10)] = 6.0 x 10-6 T 2 . BY due to X = [(4π x 10-7)(5.0)] / [2π (0.10)] = 1.0 x 10-5 T 3 . Using magnetic force, F = BILsinθ, Magnetic force on wire X, Fx = (6.0 x 10-6)(5.0)Lsin 90o = (3.0 x 10-5)L Magnetic force on wire Y, FY = (1.0 x 10-5)(3.0)Lsin 90o = (3.0 x 10-5)L Since the two wires are equal in lengths, Fx and FY are equal in magnitude. (b) (i)

Raffles Institution Year 5-6 Physics Department 2 (ii) Vertically downwards. 2 (a) (i) As seen from the increased balance reading, there is a downward force on the magnet due to current-carrying wire. By Newton’s third law, there is an upward force on the wire by the magnet. By Fleming’s left hand rule, pole P is a North pole (ii) Consider the forces on the magnet: N = BIL where N is the increase in contact force between the magnet and the top -pan balance caused by the current passing through the wire in a magnetic field (2.3 × 10–3)(9.81) = B(2.6)(4.4 × 10–2) B = 0.20 T (b) In an electric field, the force on a moving charged particle acts along the direction of the electric field while the magnetic force acts perpendicular to its motion and the magnetic field.

Raffles Institution Year 5-6 Physics Department 3 Either: In an electric field, a stationary charged particle will experience an electric force but will not experience a magnetic force in a magnetic field. Or: In an electric field, a charged particle moving parallel to the electric field will experience a force which causes it to accelerate or decelerate but a charged particle moving parallel to the magnetic field will not experience a force and will move with constant speed. 3 (a) The current in AC flows in a direction perpendicular to the solenoid’s magnetic field , which acts along the axis of the solenoid. A magnetic force will act on AC and its direction is downwards according to Fleming’s Left Hand Rule. This force will have a turning effect about the pivot along XY as its line of action is some distance away. (b) The directions of the currents in AX and CY are anti -parallel and parallel respectively to that of the magnetic flux density of the solenoid. Hence no force will be induced on AX and CY. (c) By the principle of moments, taking moments about XY: anti-clockwise moment = clockwise moment ( )( ) ( )( )60 40BIL mg = B ( )( )( ) ( )( )( ) 0.030 9.81 40 3.0 0.200 60= 0.327T= 4 ( ) ( ) ( ) , whe