Topic 11 Electromagnetic Induction For IB Physics HL
- Question 1
- Question 2
- Question 3
- Question 4
- Question 5
- Question 6
- Question 7
- Question 8
- Question 9
- Question 10
Explain, with reference to electromagnetic induction, the effect of the motion of the coil on the current.
- (as the coil moves the) conductor cuts the magnetic field / there is a change in flux linkage;
- induces an emf across the coil / a current through the coil;
- opposes the driving potential difference;
- reduces the (net) current;
This question is about changing magnetic fields.
A single-turn conducting square coil is released and falls vertically from rest. At the instant it is released, the coil is at the boundary of a region of a uniform horizontal
magnetic field directed into the plane of the paper as shown. The ends of the coil are not joined together.
Each side of the coil is 0.050 m long. The dimensions of the magnetic field region are greater than that of the coil. The magnetic field strength is 25 mT.
a) Calculate the electromotive force (emf) induced in the coil at the instant just before the whole of the coil enters the magnetic field.
speed as the whole of the coil enters the field
= ∈=(-)∆φ∆t = 0.05 × 0.99 × 25 × 10-3
1.23(mV) or 1.24(mV0; (allow use of g=10 to give 1.25(mV))
b) Suggest why the time taken for the whole of the coil to enter the magnetic field increases if the coil is a continuous loop.
current (induced) in the coil;
this will act so as to oppose the movement/reference to Lenz’s law;
the force will be upwards/resistive/counteracts the effect of gravitational force;
a) State and explain the direction of the current induced in the ring during this change.
- field caused by (induced) current must be downwards;
- to oppose the change that produced it;
- hence the current must be clockwise;
b) The following data are available.
- The resistance of ring = 3.0×10-3 Ω
- Initial magnetic flux = 1.2×10-5 Wb
- Final magnetic flux = 2.4×10-5 Wb
- Time interval = 2.0×10-3s
Calculate the average current induced in the ring
or 6.0×10-3 (v);
a) A bar magnet falls vertically from rest through a coil of wire. The potential difference (pd) across the coil is recorded by a datalogger.
The graph shows the variation with time of the pd across the coil.
(i) Explain, with reference to Faraday’s and Lenz’s laws, the shape of the graph.
(ii) The coil has 1500 turns. Calculate the magnitude of the maximum rate of change of magnetic flux.
- rate of change of flux (linkage) leads to induced emf (Faraday);
- the direction of emf tends to oppose the change (Lenz);
- thus emf in one direction as magnet enters and in the opposite direction as it leaves coil;
- magnet going faster so second peak larger;
- magnet going faster so the width of the second peak is less;
(ii) attempted use of
recognition that the maximum pd is 0.8 (V);
5 x 3 x 10-4 (Wbs-1)
b) The magnet is now suspended from a spring. The magnet is displaced vertically and starts to oscillate in and out of the coil. A sinusoidal alternating current of RMS value 280 nA is induced in the coil.
(i) State in words how the RMS value of the alternating current relates to a direct current of 280nA.
(ii) The coil has a resistance of 1.5MΩ. Calculate the peak voltage across the coil.
(iii) Explain what effect the generation of the current has on the oscillation of the magnet.
(i) the value of the direct current (or voltage) that dissipates same power (in a resistor);
V =I R=0.59 (V);
(iii) damps oscillation / OWTTE; dissipation of energy in coil/magnet;
a) loop of copper wire in a region of a uniform magnetic field is rotated about a horizontal axis.
The magnitude of the magnetic field strength is B and the area of the loop is A.
(i) State the minimum value and the maximum value of the magnetic flux linking the loop.
(ii) Outline with reference to Faraday’s law why, if the speed of rotation of the loop is increased, the maximum emf induced in the loop is increased.
(i) minimum: zero / –BA (minus sign required) maximum: BA
(ii) Main points:
- (Faraday’s law states that the) induced emf equals/is proportional to the rate of change of flux/flux linkage;
- speed greater so time for change shorter / flux (linkage) is unchanged;
- greater rate of change (of flux etc) gives a greater (induced) emf;
b) The loop in (a) is connected in series with a resistor of resistance 15 Ω. The root mean squared (RMS) value of the sinusoidal current in the resistor is 2.3 mA.
(i) Explain what is meant by the RMS value of sinusoidal current.
(ii) Determine the maximum power dissipated in the resistor.
(i) (equivalent) direct current; dissipating same power in a (fixed) resistor (as the RMS current);
(ii) maximum current =;
maximum power = mW;
a) Define magnetic flux
the product of (the magnitude of) the normal component of magnetic field strength;
and area through which it passes/with which it is associated;
φ = BA cos θ ;
all terms defined/shown on a diagram;
b) A coil is rotated at a constant speed in a region of a uniform magnetic field. The graph shows the variation with time t of the emf ε induced in the coil for one cycle of rotation.
(i) On the graph label, with the letter T, a time at which the flux linkage in the coil is a maximum.
(ii) Use the graph to determine the rate of change of flux at t=4.0ms. Explain your answer.
(iii) Calculate the root mean square value of the induced emf.
(i) letter T clearly marked at 5.0ms or 15ms;
(ii) 2.0V/Wbs; emf equals the rate of change of flux; The use of slope to obtain answer is incorrect – this yields a value of 1.8.
a) A rod made of conducting material is in a region of a uniform magnetic field. It is moved horizontally along two parallel conducting rails X and Y. The other ends of the rails are connected by a thin conducting wire.
The speed of the rod is constant and is also at right angles to the direction of the uniform magnetic field.
(i) Describe, with reference to the forces acting on the conduction electrons in the rod, how an emf is induced in the rod.
(ii) An induced emf is produced by a rate of change of flux. State what is meant by a rate of change of flux in this situation.
(i) electrons are moving at right angles to the magnetic field;
electrons experience a force directed along the rod/charge is separated in the rod;
the work done by this force to achieve this separation leads to an induced emf;
(ii) the product of the magnitude of field strength and the rate at which the area is swept out by the rod is changing / the rate at which the rod cuts through field lines;
b) The length of the rod in (a) is 1.2 m and its speed is 6.2 ms. The induced emf is 15 mV.
(i) Determine the magnitude of the magnetic field strength through which the rod is moving.
(ii) Explain how Lenz’s law relates to the situation described in (a).
(i) B = ;
(ii) Lenz's law states that the direction of the induced emf/current is such as to oppose the change producing it;
There is a current in the rod due to the induced emf;
The force on the current/rod due to the magnetic field is in the opposite direction to the force producing the motion of the rod;
a) State Lenz's law
Induced emf/induced current acts so as to oppose the change causing it;
b) Two identical aluminium balls are dropped simultaneously from the same height. Ball P falls through a region with no magnetic field. Ball Q falls through a region of uniform horizontal magnetic flux density B.
ball Q enters/leaves the magnetic field/experiences changing flux;
so an emf/current is induced;
this causes a magnetic field;
which opposes the motion of / exerts an upward force on ball Q;
In terms of energy:
ball Q moves through a magnetic field/experiences changing flux;
so an emf/current is induced;
current causes dissipative heating due to resistance;
some kinetic energy changes to thermal energy;
The diagram shows a horizontal metal rod suspended by two vertical insulated springs.
The rod moves vertically up and down with simple harmonic motion with a time period T at right angles to a uniform magnetic field. The diagram shows the variation with time t of the vertical displacement x of the rod.
a) On the axes provided, draw a graph to show
(i) the variation with time t of the vertical velocity v of the rod.
(ii) the variation with time t of the emf generated between the ends of the rod.
cosine wave same frequency as the original; phase correct;
(ii) emf in phase or antiphase with the answer to (a)(i);
b) The length of the rod is 0.18 m and the magnitude of the magnetic field is 58 μT. The frequency of the simple harmonic motion is 2.5 Hz. The amplitude of the motion is 8.2×10 m. Determine the magnitude of the maximum emf ε between the ends of the rod.
max speed =8.2×10- 2 ×2π×2.5(=1.29ms-1 );
c) The frequency of the motion is doubled without any change in the amplitude of the motion. State and explain the changes to the variation with time t of the emf ε generated as a result of this change in frequency.
frequency of the emf doubles/period halves;
because of the same change of flux in half the time / because the frequency of emf must equal frequency of oscillation;
maximum emf doubles;
maximum speed doubles/flux changes at twice rate;
a) The electron’s path while in the region of a magnetic field is a quarter circle. Show that the time the electron spends in the region of the magnetic field is 7.5×10-11 s.
= 7.5×10-11 s
b) A square loop of conducting wire is placed near a straight wire carrying a constant current I. The wire is in the same plane as the loop.
The loop is made to move with constant speed v towards the wire.
(i) Explain, by reference to Faraday’s and Lenz’s laws of electromagnetic induction, why work must be done on the loop.
(ii) Suggest what becomes of the work done on the loop.
(i) the flux in the loop is changing and so (by Faraday’s law) an emf will be induced in the loop;
(by Lenz’s law) the induced current will be (counter-clockwise) and so there will be a magnetic force opposing the motion;
requiring work to be done on the loop;
(ii) it is dissipated as thermal energy (due to the resistance) in the loop/radiation;