IB Physics HL Theme D: Fields questions

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Question 1

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)
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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 2

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
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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 3

Multiple choiceHard1 mark

A straight horizontal conductor of length 0.50 m0.50\ \mathrm{m} carries a current of 2.0 A2.0\ \mathrm{A} northeast; its eastward:northward direction-component ratio is 3:43:4 A uniform horizontal magnetic field of magnitude 0.50 T0.50\ \mathrm{T} points northeast, with eastward:northward direction-component ratio 4:34:3 What are the magnitude and direction of the force on the conductor?

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Answer: C. The current and field directions are 53.1∘53.1^\circ and 36.9∘36.9^\circ north of east, so the angle between them is 16.2∘16.2^\circ with sin⁡θ=0.28\sin\theta=0.28 Thus F=BILsin⁡θ=(0.50)(2.0)(0.50)(0.28)=0.14 NF=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⁡θ\cos\theta giving 0.48 N0.48\ \mathrm{N} and reverses the force direction.
D: uses cos⁡θ\cos\theta giving 0.48 N0.48\ \mathrm{N} but gives the correct direction.
Remember: the force depends on the component of the current direction perpendicular to the magnetic field.

Question 4

Multiple choiceHard1 mark

A circular loop of radius rr is in a uniform magnetic field of magnitude BB The angle between the field and the loop’s normal is 60∘60^\circ The radius is changed to 2r2r the field magnitude to B/2B/2 and the angle to 30∘30^\circ What is the ratio of the final magnetic-flux magnitude to the initial magnetic-flux magnitude?

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Answer: C. The magnetic-flux magnitude is Φ=BAcos⁡θ\Phi=BA\cos\theta and the loop’s area is proportional to r2r^2 Thus

ΦfinalΦinitial=12×4×cos⁡30∘cos⁡60∘=23.\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 1/21/2 from the reduced field magnitude.
Remember: the angle in the flux expression is measured from the loop’s normal.

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