IB DP Physics D. 3 Motion in electromagnetic fields IB Style Question Bank HL Paper 2


An electrically heated pad is designed to keep a pet warm.
The pad is heated using a resistor that is placed inside the pad. The dimensions of the resistor are shown on the diagram. The resistor has a resistance of \(4.2 \Omega\) and a total length of \(1.25 \mathrm{~m}\).

When there is a current in the resistor, the temperature in the pad changes from a room temperature of \(20^{\circ} \mathrm{C}\) to its operating temperature at \(35^{\circ} \mathrm{C}\).
(a) The designers state that the energy transferred by the resistor every second is \(15 \mathrm{~J}\).
Calculate the current in the resistor. [1]

(b) The designers wish to make the resistor from carbon fibre.
The graph shows the variation with temperature, in Kelvin, of the resistivity of carbon fibre.

(ii) The power supply to the pad has a negligible internal resistance.
State and explain the variation in current in the resistor as the temperature of the pad increases.[2]

(c) When there is a current in the resistor, magnetic forces act between the resistor strips.
For the part of the resistor labelled RS,

(i) outline the magnetic force acting on it due to the current in \(P Q\).[1]

(ii) state and explain the net magnetic force acting on it due to the currents in \(P Q\) and \(T U\).[2]

(d) The design of the pad encloses the resistor in a material that traps air. The design also places the resistor close to the top surface of the pad.

Explain, with reference to thermal energy transfer, why the pad is designed in this way. [3]



a $
I=\alpha \sqrt{\frac{P}{R}}=» 1.9 \ll A » \checkmark
ALTERNATIVE 1 (Calculation of length)
Read off from graph [ \(\left.2.8-3.2 \times 10^{-5} \Omega \mathrm{m}\right]^{\checkmark}\)
Use of \(I=\frac{R A}{\rho}\)
I=1.3-1.4 \ll \mathrm{m} \otimes \checkmark
ALTERNATIVE 2 (Calculation of area)
Read off from graph [ \(\left.2.8-3.2 \times 10^{-5} \Omega \mathrm{m}\right]^{\checkmark}\)
Use of \(A=\rho \frac{I}{R} \checkmark\)
A=8.3-9.5 \times 10^{-6} \alpha^2 » \checkmark
ALTERNATIVE 3 (Calculation of resistance)
Read off from graph [ \(\left.2.8-3.2 \times 10^{-5} \Omega \mathrm{m}\right]^{\vee}\)
Use of \(R=\rho \frac{1}{A} \checkmark\)
R=3.6-4.2 * \Omega » \checkmark
ALTERNATIVE 4 (Calculation of resistivity)
Use of \(\rho=\frac{R A}{I}\)
\rho=3.2 \times 10^{-5} \ll \Omega \mathrm{m} * \checkmark
Read off from graph \(260-280 \mathrm{~K} \checkmark\)

b ii «Resistivity and hencew resistance will decrease
«Pd across pad will not change because internal resistance is negligiblew Current will increase \(\checkmark\)

c i «The force is” away from \(\mathrm{PQ} /\) repulsive/to the right

c iiThe magnetic fields «due to currents in \(P Q\) and \(T U_n\) are in opposite directions \(O R\)
There are two «repulsiven forces in opposite directions \(\checkmark\)
Net force is zero \(\checkmark\)

d Air is a poor «thermal» conductor \(\checkmark\)
Lack of convection due to air not being able to move in material
Appropriate statement about energy transfer between the pet, the resistor and surroundings

The rate of thermal energy transfer to the top surface is greater than the bottom “due to thinner material»


A proton moves along a circular path in a region of a uniform magnetic field. The magnetic field is directed into the plane of the page.

    1. Label with arrows on the diagram the

      1. magnetic force F on the proton. [1]

      2. velocity vector v of the proton. [1]

    2. The speed of the proton is 2.16 × 106 m s1 and the magnetic field strength is 0.042 T. For this proton,

      1. determine, in m, the radius of the circular path. Give your answer to an appropriate number of significant figures. [3]

      2. calculate, in s, the time for one full revolution. [2]



a i F towards centre

a ii

v tangent to circle and in the direction shown in the diagram 

b i

«qvB = \(\frac{mv^2}{R}\) ⇒ » R = \(\frac{mv}{qB}\)/ \(\frac{1.673\times 10^{-27}\times 2.16\times 10^6}{1.60\times 10^{-19}\times 0.042}\)

R = 0.538 «m»
R = 0.54 « m»

b ii

T = \(\frac{2πR}{v}\)/ \(\frac{2π×0.54}{2.61×10^6}\)

T 1.6 ×10 «s»

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