DHS 17 Electromagnetic Induction (Tutorial Answers)

Dunman High School (Senior High) H2 Physics (Syllabus 9749) Topic 17: Electromagnetic Induction Y6 2025 Tutorial Solutions 1 For Internal Use Only 1 During the rotation,  there is no change in magnetic flux density B passing through the coil  the change in total magnetic flux is given by:   ocos cos0 cos 1 , which is negative decreases f i BA BA BA           Answer: C 2 magnetic flux through the loop of wire:    6 4 2 o 8cos 65 10 T 12 10 m cos30 6.8 10BA          Answer: B B θ normal A normal θ

Dunman High School (Senior High) H2 Physics (Syllabus 9749) Topic 17: Electromagnetic Induction Y6 2025 Tutorial Solutions 2 For Internal Use Only 14 April 2025 3 magnetic flux density, B magnetic flux,  = BA = B × (8.0 × 10−3) magnetic flux linkage, N = 800  = 6.4B magnetic flux density / T magnetic flux / Wb magnetic flux linkage / Wb turns A 3.4 2.7 × 102  220  22 B 2.7 × 102 3.4  2.2 × 104 2700  0.173 C 3.4 × 104 4.2 × 102  2.7 × 106 34  2.2 × 103 D 4.2 × 102 3.4 × 104  0.27  Answer: D 4 B Since speed is constant, the change of flux linkage in entering and leaving the solenoid would take place over the same period of time, hence induced e.m.f. have the same maximum value. (Faraday’s Law) normal to area of coils of solenoid When the whole magnet remains well within the long solenoid, with the flux linkage being the maximum and remaining unchanged, the induced e.m.f. remains zero. (Faraday’s law) When magnet is entering and leaving the solenoid, the total change in flux linkage (area under the graph) is the same. 2 1 area under graph t t dN dt N dt t           N 

Dunman High School (Senior High) H2 Physics (Syllabus 9749) Topic 17: Electromagnetic Induction Y6 2025 Tutorial Solutions 3 For Internal Use Only 14 April 2025 i. first, the switch is closed ii. then the variable resistor in series with the cell is decreased iii. then the circuit containing resistor R is moved to the left iv. then the switch is opened. 5 D T/2 Since ߝ= − ௗேః ௗ௧ = − ௗ(ே஻஺) ௗ௧ = −ܣܰ ௗ஻ ௗ௧ , induced e.m.f is the largest when the gradient of the graph is the steepest.    0 0 0 cossin cos 2Note: 2 d B tdN dB NA NA NAB tdt dt dt f T t                  6 Direction of induced current in loop A: clockwise B: no induced current C: counter-clockwise Explanation for direction of induced current in loop A: The increase in current in the wire causes increasing magnetic flux linkage out of the plane of loop A. By Lenz’s law, the induced current due to the induced e.m.f. must flow to reduce the flux linkage out of the plane of the loop. Thus , clockwise current flows through loop A. 7 (a) Since I decreases, magnetic flux linkage into the plane through the coil decreases. Hence by Lenz’s law, the current due to the induced e.m.f. must flow to increase the flux linkage into the plane of the coil. Thus, clockwise current flows through the loop and flows from left to right through R. (b) I A B C current I decreases rapidly to zero A B R

Dunman High School (Senior High) H2 Physics (Syllabus 9749) Topic 17: Electromagnetic Induction Y6 2025 Tutorial Solutions 4 For Internal Use Only 14 April 2025 (i) When the switch is first closed in B, the current flowing increases from 0 to a steady value, causing increasing leftward magnetic flux linkage through the coil in A. By Lenz’s law, the induced magnetic field is rightward to oppose the change. To produce this induced field, the induced current must be clockwise as seen from the left of coil (from the fish’s point of view). Thus, the current flows from right to left through R. (ii) When the variable resistor in B is decreased, the current flowing in B increa ses, causing increasing leftward magnetic flux linkage through the coil in A. By Lenz’s law, the induced magnetic field is rightward to oppose the change. To produce this induced field, the induced current must be clockwise as seen from the left of coil. Thus, the current flows from right to left through R. (iii) Motion of the circuit A to the left causes decreasing leftward magnetic flux linkage through coil in A. B y Lenz’s law, the induced magnetic field is leftward to oppose the change. To produce this induced field, the induced current must be anticlockwise as see from the left of coil. Thus, current flows from left to right through R. (iv) When the switch is opened, the current flowing in circuit B decreases from its steady value to 0, causing decreasing leftward magnetic flux linkage through the coil in A. By Lenz’s law, the induced magnetic field is leftward to oppose the change. To produce this induced field, the induced current must be anticlockwise as seen from the left of coil. Thus, the current flows from left to right through R. 8 (a)(i) When the loop drops between the poles of the magnet from above, there is an increase in flux through the loop. From Lenz’s law, the current due to the induced e.m.f. flows in a direction such that it produces magnetic fields that oppose the increase in magnetic flux. Hence, the induced current would be clockwise to produce a magnetic field pointing opposite to the magnetic field of the magnet. (ii) When the loop drops below the poles of the magnet, there is a decrease in flux through the loop. From Lenz’s law, the current due to the induced e.m.f. flows in a direction such that it produces magnetic fields that oppose the decrease in magnetic flux. Hence, the induced current would be anti-clockwise to produce a magnetic field in the same direction as the magnetic field of the magnet. (b)(i) Considering the free body diagram of the loop of wire when it is away from the magnet: At constant speed, T = W When the loop is below the magnet, there is a increase in magnetic flux linkage through the loop as it is being pulled upwards. From Lenz’s law, the current due to Weight W Tension T

Dunman High School (Senior High) H2 Physics (Syllabus 9749) Topic 17: Electromagnetic Induction Y6 2025 Tutorial Solutions 5 For Internal Use Only 14 April 2025 the induced e.m.f. flows in a direction such that the coil experiences a downward force FB to oppose the upward motion of the loop. (The induced magnetic forces repel the magnet when the ring is approaching.) At constant speed, T = W + FB Hence the tension of the string is greater than the weight of the loop. (ii) When the loop is above the magnet, there is an decrease in magnetic flux linkage through the loop as it is bei