# IBDP Physics 5.4 –Magnetic effects of electric currents: IB Style Question Bank HL Paper 2

### Question

A proton is moving in a region of uniform magnetic field. The magnetic field is directed into the plane of the paper. The arrow shows the velocity of the proton at one instant and the dotted circle gives the path followed by the proton.

(a) Explain why the path of the proton is a circle. [2]

(b) The speed of the proton is 2.0 × 106 m s–1 and the magnetic field strength B is 0.35 T.

(i) Show that the radius of the path is about 6 cm. [2]

(ii) Calculate the time for one complete revolution.

(c) Explain why the kinetic energy of the proton is constant. [2]

Ans:

a magnetic force is to the left «at the instant shown» OR explains a rule to determine the direction of the magnetic force  force is perpendicular to velocity/«direction of» motion OR force is constant in magnitude  force is centripetal/towards the centre

b i qvB = $$\frac{Mv^2}{R}$$ R = $$\frac{1.67\times 10^{-27\times 2.0\times 10^6}}{1.6\times 10^{-19\times 0.35}}$$ OR 0.060 « m »

b ii T = $$\frac{2\pi R}{V}$$ T = « $$\frac{2\pi \times 0.06 }{2.0\times 10^5}$$ » 1.9 × 10-7 «s» 4. c ALTERNATIVE 1 work done by force is change in kinetic energy  work done is zero/force perpendicular to velocity  ALTERNATIVE 2 proton moves at constant speed  kinetic energy depends on speed

### Question

(a) The primary coil of a transformer is connected to a 110 V alternating current (ac) supply. The secondary coil of the transformer is connected to a 15 V garden lighting system that consists of 8 lamps connected in parallel. Each lamp is rated at 35 W when working at its normal brightness. Root mean square (rms) values are used throughout this question.

1. The primary coil has 3300 turns. Calculate the number of turns on the secondary coil. [1]

2. Determine the total resistance of the lamps when they are working normally. [2]

3. Calculate the current in the primary of the transformer assuming that it is ideal. [2]

4. Flux leakage is one reason why a transformer may not be ideal. Explain the effect of flux leakage on the transformer. [2]

(b) A pendulum with a metal bob comes to rest after 200 swings. The same pendulum, released from the same position, now swings at 90° to the direction of a strong magnetic field and comes to rest after 20 swings.

Explain why the pendulum comes to rest after a smaller number of swings. [4]

Ans:

a i « $\frac{15}{110}\times&space;3300$  = » 450 «turns»

a ii

ALTERNATIVE 1  calculates total current = $\frac{35}{15}\times&space;8$ « = 18.7 A»  resistance = « $\frac{15}{18.7}$ = » 0.80 «Ω»

ALTERNATIVE 2 calculates total power = 35×8 « = 280 W»  resistance =« $\frac{15^2}{280}$ =» 0.80 «Ω»

ALTERNATIVE 3 calculates individual resistance = $\frac{15^2}{35}$« =  6.43 Ω»  resistance = «$\frac{6.43}{8}$ » = 0.80 «Ω»

a iii

total power required = 280 «W»

OR uses factor $\frac{3300}{450}$

OR total current = 18.7 « A»

current = 2.5 OR 2.6 «A»

a iv

the secondary coil does not enclose all flux «lines from core»  induced emf in secondary

OR power transferred to the secondary

OR efficiency is less than expected

Award [0] for references to eddy currents/heating of the core as the reason. Award MP2 if no reason stated.

b

bob cuts mag field lines

OR there is a change in flux linkage

induced emf across bob  leading to eddy/induced current in bob

eddy/induced current produces a magnetic field that opposes «direction of» motion

force due to the induced magnetic field decelerates bob  damping of pendulum increases/there is additional «magnetic»

damping  MP4 and MP5 can be expressed in terms of energy transfer from kinetic energy of bob to electrical/thermal energy in bob