17 EMI 2022 - Tutorial ans with assignment

St. Andrew’s Junior College H2 Physics 17-33 TUTORIAL 17: EMI SOLUTIONS Level 1 Solutions 1 B. Refer to Example 2. [1] 2(a) Magnetic flux is defined as the product of the magnetic flux density and the area normal to the field through which the field is passing. Magnetic flux linkage in a coil is defined as the product of the magnetic flux passing through a coil and the number of turns of the coil. [1] [1] 2(b) The magnetic flux density is defined as the force acting per unit current in a wire of unit length at right-angles to the field whereas magnetic flux is defined as the product of the magnetic flux density and the area normal to the field through which the field is passing. [1] [1] 2(c) Faraday’s Law states that magnitude of an induced emf is directly proportional to the rate of change of magnetic flux-linkage. [1] 2(d) Lenz’s Law states that the direction of the induced emf is such that its effects oppose the change which causes it. The induced emf produces effects which opposes the change in the magnetic flux linkage. Consequently, work has to be done by an external agent to overcome this opposition and energy is conserved. Hence the electrical energy associated with the induced current is derived from the work that is done by the external agent to overcome the change in magnetic flux. If the effects of the induced emf do not oppose the change in magnetic flux, there will be a gain in energy without any work done. This violates the law of conservation of energy. [1] [1] [1] [1] 3 “its axis parallel to a uniform magnetic field ” means the B field vector is perpendicular to the plane containing the cross-sectional area of the coil. EMF =  N B A/t = 120  (80m – 20m)  0.07 / 4.0 = 130 mV [1] [1]

St. Andrew’s Junior College H2 Physics 17-34 Level 2 Solutions 4 A  The current in the solenoid sets up a B-field.  “Original current” = Applied emf ÷ Tot Resistance of circuit.  The iron rod strengthens the B-field, ie increases the flux density B, ( Topic 16: EM Learning Outcome (d); B = ro nI -NOT in Syll). r:relative permeability of the medium (soft iron)  1. When the rod is “far eough” fr solenoid: equivalent to being “not there at all”: current = “original current”  2. When the iron rod is entering the solenoid: (ie while partially in solenoid), the flux linkage increases with time.This induces an emf in the solenoid. By Lenz’s law the direction of the induced emf is opposite that of the applied emf (by the battery); resulting emf decreases, causing the current in the solenoid to decrease (momentarily). (See the dip, the one that occurs earlier in time.)  3. “Deep within the solenoid”: Here B has increased to its max; it remains constant (until it “starts to leave the solenoid). Hence flux linkage is no longer changing when the rod is deep within the solenoid; no more induced emf; resulting current reverts to its original value.  4. When the iron rod is leaving the solenoid: flux linkage is decreasing. This induces an emf. By Lenz’s law the direction of the induced emf will be in the same direction as that of the applied emf; resulting emf increases, causing the current in the solenoid to increase momentarily. (See the spike up, the one that occurs later in time.)  5. When rod is completely out of the solenoid: Induced emf becomes zero; flux linkage becomes constant again; no more induced emf; resulting current reverts to original value. 5(a) [1] 5(b) By Faraday’s law, the induced EMF is equal to the rate of change of magnetic flux linkage, which is determined by the velocity of the magnet. [1] [1] [1] t / s E / V

St. Andrew’s Junior College H2 Physics 17-35 The graph is sinusoidal because the velocity of the magnet is varying sinusoidally according to the simple harmonic motion of the spring. 5(c) Explanation in terms of Energy: When a resistor is connected energy from the induced EMF is dissipated in the resistor ( in the form of Joule heating) this energy comes from the energy of oscillations. Thus the motion of the magnet becomes more damped and amplitude decreases. OR Explanation in terms of induced Force : When a resistor is connected, induced current starts to flow in the s circuit. By Lenz’s law, this induced current will produce a magnetic field to oppose the oscillations. Thus, when the bar magnet is approaching the coil, the top end of the coil will be a magnetic pole which repels the magnet. And when the magnet is receding, the induced magnetic pole of the top end of the coil will be exerting an attractive force on the magnet. Thus the motion of the magnet becomes more damped and amplitude decreases. [1] [1] [1] [1] [1] [1] 6(a)(i) 6(a)(ii) Elaboration: (i) Deduce (fr given I-t graph), I = - Io sin wt eqn (A) Since B = 0nI, eqn (B), B = 0 n (- Io sin wt), (1) Fr eqn (1) & (A),  B-t graph : Same shape as I-t graph & in phase as well. (ii) Deduce Bo = 0nIo (2) E - - d/dt = - d (NBA)/dt = - AN dB/dt where B = - Bosin wt (3) fr (1) & (2) = - AN [ -Bo w coswt ] = +( ANonIow ) cos wt Hence E-t graph: +cosine graph; & hence out of phase by 900 wrt I-t graph [1] [1]

St. Andrew’s Junior College H2 Physics 17-36 6(b)(i) Since E = +( ANonIow ) cos wt, Comparing it with the general eqn, E = Eo cos wt  the amplitude Eo = wANonIo eqn (C) The ferrous core causes the permeability of the medium to become ro where r > 1. Hence amplitude of E, ie Eo increases when the ferrous core is introduced. As for frequency of E, (the output), it depends only on the frequency of the input current. Since the input freq remains constant, the freq of E remains unchanged. 6(b)(ii) I = - Io sin ( t ), f 2 Since E = +( ANonIow ) cos wt, As frequency f of current increases , angular frequency  also increases. Hence amplitude of E ( wANonIo ) and frequency of E-t increase. [1] [1] 6(b)(iii) Increase in the amplitude of the current Io will increase the amplitude of E ; see eqn (C). Since freq of input current remains unchanged, freq of output E is also unchanged. [1] [1] 7 Induced emf = rate of change of magnetic flux linkage = t NBA = ).)(.( 08005075 t B Emf = RI = (8)(0.1) = 0.8 V ).)(.( . 08005075 80 t B = 2.67 T s – 1 [1] [1] 8 Assume the blades are rotating in a horizontal plane. Consider the area ‘swept’ by the moving rod (ie blade) A = π(r)2 B = 5.0 × 10 – 5 T f = 2 Hz Induced emf = t NBA = 5.0 )3)((100.5 25  Consider a time = 1 period,T = 1/f = 2.83 mV [1] [1] 9(a) There is no magnetic force when particle is stationary or when it is moving parallel to the magnetic field. [1] [1] 9(b)(i) The current flows in the same direction in each coil of the spring. [1] [1]

St. Andrew’s Junior College H2 Physics 17-37 Since attractive forces experienced by parallel conductors when current flows through them in the same direction, each coil of the spring experiences attractive forces due to the neighbouring coils. 9(b)(ii) The spr