IBDP Physics 11.1 – Electromagnetic induction: IB style Question Bank HL Paper 2

Question

The diagram shows an alternating current generator with a rectangular coil rotating at a constant frequency in a uniform magnetic field.

                 

(a) Explain, by reference to Faraday’s law of induction, how an electromotive force (emf) is induced in the coil.    [3]

Answer/Explanation

Ans:

there is a magnetic flux «linkage» in the coil / coil cuts magnetic field ✓
this flux «linkage» changes as the angle varies/coil rotates ✓
«Faraday’s law» connects induced emf with rate of change of flux «linkage» with time ✓

Question

Part 2 Electromagnetic induction

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.

[3]
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 8.2×10-2 m.

Determine the magnitude of the maximum emf εmax ¬between the ends of the rod.

[3]
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 \(\varepsilon \)¬ generated as a result of this change in frequency.

[4]
 
Answer/Explanation

Markscheme

a.

(i)

cosine wave same frequency as original;
phase correct;

(ii) emf in phase or antiphase with answer to (a)(i);

b.

max speed =8.2×10-2×2π×2.5(=1.29ms-1);
ε=58×10-6×0.18×1.29;
13.5 μV;

c.

frequency of the emf doubles / period halves;
because same change of flux in half the time / because frequency of emf must equal frequency of oscillation;
maximum emf doubles;
maximum speed doubles / flux changes at twice rate;

Question

This question is about induced electromotive force (emf).

A loop of copper wire in a region of uniform magnetic field is rotated about a horizontal axis.

a.

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.

[4]
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 a sinusoidal current.

(ii) Determine the maximum power dissipated in the resistor.

[4]
 
Answer/Explanation

Markscheme

a.

(i) minimum: zero / –BA (minus sign required)
maximum: BA } (both needed)

(ii) Look for these main points:
(Faraday’s law states that the) induced emf equals/is proportional to the rate of change of flux/flux linkage; { (must see induced)
speed greater so time for change shorter / flux (linkage) is unchanged;
greater rate of change (of flux etc) gives a greater (induced) emf;
Award [1 max] if answer states flux (linkage) change is larger.

b.

(i) (equivalent) direct current;
dissipating same power in a (fixed) resistor (as the rms current);

(ii) maximum current \( = \left( {\sqrt 2 \times 2.3 = } \right)3.2/3.3{\rm{mA}}\);
maximum power = (3.32×15=)0.16mW;