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.

Question 1

Paper 2Easy6 marks

A probe is outside a small spherical moon. Figure 1 shows the gravitational potential VgV_g as a function of distance rr 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

gr=−ΔVgΔr.g_r=-\frac{\Delta V_g}{\Delta r}.
1.51.61.71.81.922.12.22.32.42.5-2.8-2.6-2.4-2.2-2-1.8-1.6-1.4r / (106 m)Vg / (106 J kg-1)
(a)

Read the graph and state VgV_g at r=2.0×106 mr=2.0\times10^6\ \mathrm{m} and r=2.5×106 mr=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=2.0×107 mr=2.0\times10^7\text{ m} from the centre of a planet. The graph shows the planet’s gravitational field strength gg at selected distances rr What is the escape speed from this position?

0123405101520r/(107 m)g/(N kg-1)
Show answer and explanation

Answer: B. At r=2.0×107 mr=2.0\times10^7\text{ m} the graph gives g=5.0 N kg−1g=5.0\text{ N kg}^{-1} Since g=GM/r2g=GM/r^2 vesc=2GM/r=2gr=2(5.0)(2.0×107)=1.4×104 m s−1v_{\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 gr\sqrt{gr} omitting the factor 2 under the square root.
C: assumes escape speed is twice the circular orbital speed, rather than 2\sqrt{2} times it.
D: reads g=20 N kg−1g=20\text{ N kg}^{-1} from the graph at r=1.0×107 mr=1.0\times10^7\text{ m} not the value at the satellite’s radius.

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

Question 3

Multiple choiceMedium1 mark

Two fixed point charges, +Q+Q and −Q-Q are separated by 2a2a where Q>0Q>0 Point PP is midway between them. Let kk be Coulomb’s constant. Which row gives the electric potential at PP and the location chosen as the zero of electric potential?

Electric potential at PPZero-potential reference
A.00infinity
B.2kQa\dfrac{2kQ}{a}infinity
C.00PP
D.kQa\dfrac{kQ}{a}infinity
Show answer and explanation

Answer: A. For a point charge, V=kq/rV=kq/r when the potential is zero at infinity. At PP the contributions are +kQ/a+kQ/a and −kQ/a-kQ/a so their scalar sum is V(P)=0V(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+Q but omits the equal contribution from −Q-Q

Remember: electric potentials add as signed scalars.

Question 4

Paper 2Medium7 marks

A straight wire segment of length 0.150 m0.150\ \mathrm{m} carrying a current of 2.0 A2.0\ \mathrm{A} is placed in a uniform magnetic field. The angle θ\theta between the current and the magnetic field is varied, and the magnitude of the force on the wire is measured.

00.10.20.30.40.50.60.70.80.9102468101214161820222426sinθF / mN
(a)

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

[2]
(b)

A separate straight wire segment of length 0.12 m0.12\ \mathrm{m} carries a current of 1.5 A1.5\ \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 A4.0\ \mathrm{A} and 6.0 A6.0\ \mathrm{A} in the same direction. Their separation is 0.050 m0.050\ \mathrm{m} Calculate the force per unit length between the wires and state whether the force is attractive or repulsive.

Use μ0=4π×10−7 N A−2\mu_0=4\pi\times10^{-7}\ \mathrm{N\,A^{-2}}

[3]

Question 5

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.

uniform magnetic field,B1=0.040 T, into the pageuniform magnetic field,B2=0.50 T, into the page××××××××+240 V0 Vproton beam1.00 cm××××××××××××××××××××PQCr=1.25 cm
(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 1.501.50 times its original value. Using your result from (c), determine the new radius of their path in region 2.

[2]

Question 6

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.

00.10.20.30.40.5-16-12-8-40481216t / sε / VPQ
(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 Hz3.0\ \mathrm{Hz}

[2]

Question 7

Paper 2Medium8 marks

A coil has 120 turns and an area of 4.0×10−3 m24.0\times10^{-3}\,\text{m}^2 It is in a uniform magnetic field of 0.20 T0.20\,\text{T} The angle θ\theta is measured between the field and the normal to the coil.

Initially, θ=60∘\theta=60^\circ The coil rotates steadily until θ=0∘\theta=0^\circ taking 0.10 s0.10\,\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 θ=60∘\theta=60^\circ to θ=0∘\theta=0^\circ is now completed in 0.25 s0.25\,\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 T0.20\,\text{T} to 0.12 T0.12\,\text{T} in 0.050 s0.050\,\text{s} Determine the magnitude of the average induced emf.

[2]

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