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1. 1A (a) The force acting on AB points downwards. 1A The force acting on CD points upwards. (b) This is because the force acting on BC is parallel to the rotation axis. 1A (c) 1A The coil rotates in anticlockwise direction. (d) X is a commutator. 1A It changes contact from one brush to the other every half turn of the coil. 1A This reverses the current through the coil, so do the forces acting on the coil. 1A The coil, therefore, carries on rotating in the same direction. 3 1A (e) Any three of the following: Use stronger magnets. Increase the number of turns in the coil. Increase the area of the coil within the magnetic field. Increase the current. 2 (i) N (ii) S (Correct shape of magnetic field lines) 1A (Correct direction of magnetic field lines) 1A (Correct polarities) 1A An upward force acts on AB and 1A a downward force acts on CD. 1A (iii) The coil rotates in clockwise direction as seen by the observer. (iv) It changes contact from one brush to the other every half turn of the coil. 1A (Cancelled) This reverses the current through the coil, so do the forces acting on the coil. The coil, therefore, carries on rotating in the same direction. (v) The rotation direction of the motor remains unchanged. 1A It is because both the current direction and the polarities of the solenoids are reversed in reversing the battery. 1A 3 Solutions (a) Since the electric force balances the magnetic force and E = Marks VH , d we have QE = BQv. v= 1M 1 10 6 E VH = = = 5 10–5 m s–1 B Bd (2)(0.01) 1A The drift velocity of the charge carriers is 5 10–5 m s–1. (b) By VH = n= BI , nQt 1M (2)(3) BI = = 7.5 1027 m–3 5 VH Qt (5 10 )(1.6 10 19 )(0.005) 1A There are 1.5 1026 charge carriers per unit volume. (c) Since the pointer of the centre-zero galvanometer deflects to the right, the current flows to the right. 1A By Fleming’s left-hand rule, the charge carriers are deflected upwards. 1A 4 Solutions Marks (a) Downwards 1A (b) Upper side 1A (c) Since the electric force balances the magnetic force and E = we have QE = BQv. v= 4 10 6 E VH = = = 0.392 m s–1 B Bd (1.2 10 3 )(0.85 10 2 ) VH , d 1M 1A