IB Physics HL Theme D: Fields questions

Question 1

Paper 2Easy6 marks

A probe is outside a small spherical moon. Figure 1 shows the gravitational potential $V_g$ as a function of distance $r$ from the moon’s centre. The positive radial direction is defined to be away from the moon. The radial component of gravitational field strength is related to the potential gradient by

g_r=-\frac{\Delta V_g}{\Delta r}.

(a)

Read the graph and state $V_g$ at $r=2.0\times10^6\ \mathrm{m}$ and $r=2.5\times10^6\ \mathrm{m}$

[2]

(b)

Calculate the magnitude of the average radial gravitational field strength between these two distances.

[3]

(c)

State the direction of the gravitational field relative to the moon.

[1]

Question 2

Multiple choiceMedium1 mark

A satellite is at a distance $r=2.0\times10^7\text{ m}$ from the centre of a planet. The graph shows the planet’s gravitational field strength $g$ at selected distances $r$ What is the escape speed from this position?

Show answer and explanation

Answer: B. At $r=2.0\times10^7\text{ m}$ the graph gives $g=5.0\text{ N kg}^{-1}$ Since $g=GM/r^2$ $v_{\rm esc}=\sqrt{2GM/r}=\sqrt{2gr}=\sqrt{2(5.0)(2.0\times10^7)}=1.4\times10^4\text{ m s}^{-1}$

A: uses the circular orbital speed $\sqrt{gr}$ omitting the factor 2 under the square root.
C: assumes escape speed is twice the circular orbital speed, rather than $\sqrt{2}$ times it.
D: reads $g=20\text{ N kg}^{-1}$ from the graph at $r=1.0\times10^7\text{ m}$ not the value at the satellite’s radius.

Remember: escape speed is $\sqrt{2}$ times circular orbital speed at the same radius.

Question 3

Multiple choiceMedium1 mark

Two fixed point charges, $+Q$ and $-Q$ are separated by $2a$ where $Q>0$ Point $P$ is midway between them. Let $k$ be Coulomb’s constant. Which row gives the electric potential at $P$ and the location chosen as the zero of electric potential?

	Electric potential at $P$	Zero-potential reference
A.	$0$	infinity
B.	$\dfrac{2kQ}{a}$	infinity
C.	$0$	$P$
D.	$\dfrac{kQ}{a}$	infinity

Show answer and explanation

Answer: A. For a point charge, $V=kq/r$ when the potential is zero at infinity. At $P$ the contributions are $+kQ/a$ and $-kQ/a$ so their scalar sum is $V(P)=0$ The reference location for zero potential is infinity.

B: adds the magnitudes of the two contributions, ignoring the negative charge’s sign.
C: confuses a point where the net potential is zero with the location chosen as the zero reference.
D: includes the contribution from $+Q$ but omits the equal contribution from $-Q$

Remember: electric potentials add as signed scalars.

Question 4

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 5

Paper 2Medium7 marks

A straight wire segment of length $0.150\ \mathrm{m}$ carrying a current of $2.0\ \mathrm{A}$ is placed in a uniform magnetic field. The angle $\theta$ between the current and the magnetic field is varied, and the magnitude of the force on the wire is measured.

(a)

Use the gradient of Figure 1 to determine the magnetic field strength.

[2]

(b)

A separate straight wire segment of length $0.12\ \mathrm{m}$ carries a current of $1.5\ \mathrm{A}$ due east in a uniform magnetic field directed vertically upwards. The wire is perpendicular to the magnetic field. Use your answer to (a) to calculate the magnitude of the force on the wire and state its direction.

[2]

(c)

Two long parallel wires carry currents of $4.0\ \mathrm{A}$ and $6.0\ \mathrm{A}$ in the same direction. Their separation is $0.050\ \mathrm{m}$ Calculate the force per unit length between the wires and state whether the force is attractive or repulsive.

Use $\mu_0=4\pi\times10^{-7}\ \mathrm{N\,A^{-2}}$

[3]

Question 6

Paper 2Medium12 marks

A beam of protons passes through a velocity selector and then enters a region containing a uniform magnetic field. Figure 1 shows the arrangement. The diagrams are schematic and not to scale. The protons pass undeflected through the selector.

(a)

Calculate the magnitude of the electric field between the plates in region 1.

[2]

(b)

Consider the motion of the protons in region 1.

(i)

If the magnetic field in region 1 is switched off, state the direction of the acceleration of the protons while they are between the plates.

[1]

(ii)

Determine the speed of the protons that pass undeflected through the selector.

[2]

(c)

Use the path in region 2 to determine the charge-to-mass ratio of a proton.

[3]

(d)

Explain why the kinetic energy of a proton remains constant as it moves through region 2.

[2]

(e)

The speed of the protons entering region 2 is increased to $1.50$ times its original value. Using your result from (c), determine the new radius of their path in region 2.

[2]

Question 7

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 8

Paper 2Medium8 marks

A coil rotates at constant speed in a uniform magnetic field. The same coil and field are used for both rotations; only the rotational frequency changes. Figure 1 shows the induced emf for the two rotations.

(a)

Determine the rotational frequency corresponding to each trace.

[2]

(b)

Determine the ratio of the peak induced emf in Q to that in P. Using your answer to (a), deduce the relationship between peak induced emf and rotational frequency.

[2]

(c)

Explain why increasing the rotational frequency increases the peak induced emf.

[2]

(d)

Using the relationship inferred in (b), determine the peak induced emf when the same coil rotates at $3.0\ \mathrm{Hz}$

[2]

Question 9

Paper 2Medium8 marks

A coil has 120 turns and an area of $4.0\times10^{-3}\,\text{m}^2$ It is in a uniform magnetic field of $0.20\,\text{T}$ The angle $\theta$ is measured between the field and the normal to the coil.

Initially, $\theta=60^\circ$ The coil rotates steadily until $\theta=0^\circ$ taking $0.10\,\text{s}$ The magnetic field remains constant during this rotation.

(a)

Calculate the magnetic flux through one turn of the coil initially.

[1]

(b)

Determine the magnitude of the average induced emf during the rotation.

[3]

(c)

The same rotation from $\theta=60^\circ$ to $\theta=0^\circ$ is now completed in $0.25\,\text{s}$ Calculate the magnitude of the average induced emf and explain how it compares with your answer to (b).

[2]

(d)

The coil is held stationary with its normal parallel to the field. The magnetic field decreases uniformly from $0.20\,\text{T}$ to $0.12\,\text{T}$ in $0.050\,\text{s}$ Determine the magnitude of the average induced emf.

[2]

Question 10

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.

IB Physics HL Theme D: Fields questions

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