IB Physics HL Theme D: Fields questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

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Multiple choice Written

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Question 1

Paper 2Hard10 marks

In a simplified Millikan oil-drop experiment, negatively charged drops are held stationary between parallel plates. The polarity is chosen so that the electric force on a drop is upward. Assume buoyancy and other forces are negligible. The potential difference between the plates is $400\ \mathrm{V}$ and their separation is $4.00\ \mathrm{mm}$

Drop A has mass $4.90\times10^{-15}\ \mathrm{kg}$ Further drops are measured using the same method. The uncertainty in each charge estimate, including that for drop A, is $\pm0.08\times10^{-19}\ \mathrm{C}$ An independently calibrated candidate value is $e_0=1.60\times10^{-19}\ \mathrm{C}$ treat its uncertainty as negligible.

Drop	Measured charge magnitude / $10^{-19}\ \mathrm{C}$
B	3.18
C	4.79
D	6.39
E	8.01
F	5.55

(a)

Show that the magnitude $q$ of the charge on a stationary drop is given by

q=\frac{mgd}{V},

where $m$ is the drop mass, $g$ is gravitational field strength, $d$ is the plate separation and $V$ is the potential difference.

[2]

(b)

Calculate the magnitude of the charge on drop A. Give your answer in coulombs.

[2]

(c)

Using $e_0$ and your answer to (b), determine the nearest integer $n$ for each of drops A–E, where $q\approx ne_0$

[2]

(d)

A student claims that the results for drops A–E establish that $e_0$ is the smallest possible charge, while drop F disproves charge quantization. Evaluate this claim using the measurements and their uncertainties.

[4]

Question 2

Multiple choiceHard1 mark

A straight horizontal conductor of length $0.50\ \mathrm{m}$ carries a current of $2.0\ \mathrm{A}$ northeast; its eastward:northward direction-component ratio is $3:4$ A uniform horizontal magnetic field of magnitude $0.50\ \mathrm{T}$ points northeast, with eastward:northward direction-component ratio $4:3$ What are the magnitude and direction of the force on the conductor?

Show answer and explanation

Answer: C. The current and field directions are $53.1^\circ$ and $36.9^\circ$ north of east, so the angle between them is $16.2^\circ$ with $\sin\theta=0.28$ Thus $F=BIL\sin\theta=(0.50)(2.0)(0.50)(0.28)=0.14\ \mathrm{N}$ Fleming’s left-hand rule gives a downward force.

A: gives the correct magnitude but reverses the force direction.
B: uses $\cos\theta$ giving $0.48\ \mathrm{N}$ and reverses the force direction.
D: uses $\cos\theta$ giving $0.48\ \mathrm{N}$ but gives the correct direction.
Remember: the force depends on the component of the current direction perpendicular to the magnetic field.

Question 3

Multiple choiceHard1 mark

A circular loop of radius $r$ is in a uniform magnetic field of magnitude $B$ The angle between the field and the loop’s normal is $60^\circ$ The radius is changed to $2r$ the field magnitude to $B/2$ and the angle to $30^\circ$ What is the ratio of the final magnetic-flux magnitude to the initial magnetic-flux magnitude?

Show answer and explanation

Answer: C. The magnetic-flux magnitude is $\Phi=BA\cos\theta$ and the loop’s area is proportional to $r^2$ Thus

\frac{\Phi_{\mathrm{final}}}{\Phi_{\mathrm{initial}}} =\frac{1}{2}\times 4\times\frac{\cos30^\circ}{\cos60^\circ} =2\sqrt{3}.

A: reverses the ratio of the cosine factors.
B: uses the radius ratio as the area ratio, rather than squaring it.
D: omits the factor of $1/2$ from the reduced field magnitude.
Remember: the angle in the flux expression is measured from the loop’s normal.

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