IB Physics SL 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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Question 1

Multiple choiceHard1 mark

A satellite moves in an elliptical orbit around a planet. Its closest and farthest distances from the planet are $1.0\times10^7\text{ m}$ and $7.0\times10^7\text{ m}$ A satellite in a circular orbit around the same planet has radius $2.0\times10^7\text{ m}$ and period $4.0\text{ d}$ Which pair gives the elliptical orbit’s period, followed by the ratio of its speed at closest approach to its speed at farthest distance?

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Answer: C. For the same planet, Kepler’s third law gives $T^2/a^3=\text{constant}$ and Kepler’s second law gives equal swept areas in equal times.

The semi-major axis is $a=(1.0+7.0)\times10^7/2=4.0\times10^7\text{ m}$ Thus $T=4.0(4.0/2.0)^{3/2}=11.3\text{ d}$ which rounds to $11\text{ d}$
At closest and farthest approach, velocity is perpendicular to the radius, so equal area rates give $r_{\rm close}v_{\rm close}=r_{\rm far}v_{\rm far}$ Hence $v_{\rm close}/v_{\rm far}=7$ A: omits the cube in Kepler’s third-law relation, using $T/T_0=\sqrt{a/a_0}$ B: reverses the speed ratio. D: uses $T/T_0=(a/a_0)^3$ omitting the square root. Remember: orbital speed is greater nearer the central body.

Question 2

Multiple choiceHard1 mark

Two oil drops P and R have initial charges $+18Q$ and $-10Q$ where $Q>0$ They exchange charge in isolation, and afterward P has charge $+14Q$ Treat all readings as exact; $e$ is the greatest positive unit of which every initial and final charge is an integer multiple. Which pair gives $q_{R,\mathrm f}$ and the number of elementary-charge units transferred? [1]

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Answer: A [1]. The total charge of the isolated pair is conserved, and each drop’s charge is an integer multiple of $e$ The initial total is $(18-10)Q=+8Q$ so $q_{R,\mathrm f}=+8Q-14Q=-6Q$ The charge magnitudes $18Q, 10Q, 14Q$ and $6Q$ have greatest common unit $e=2Q$ P’s charge changes by $4Q$ so $4Q/(2Q)=2$ elementary-charge units are transferred.

B: uses $Q$ rather than the greatest common unit $2Q$ as $e$
C: gives the residual charge the wrong sign.
D: assigns R the same signed change as P, rather than the opposite change required by conservation.

Remember: in an isolated pair, the signed charge changes sum to zero.

Total: 1 mark.

Question 3

Multiple choiceMedium1 mark

Particles X and Y, each with charge magnitude $q$ enter the same uniform magnetic field at right angles to the field with equal speeds. Only magnetic forces act. Each follows a semicircular path, and Y’s path radius is twice X’s. Which row gives the ratios $(|q|/m)_Y/(|q|/m)_X$ and $K_Y/K_X$ respectively, after both particles complete their semicircles?

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Answer: B. The magnetic force supplies the centripetal force, so $|q|vB=mv^2/r$ giving $|q|/m=v/(Br)$ With equal speeds and the same field, the charge-to-mass ratio is inversely proportional to radius, so the ratio for Y and X is $1/2$ Equal charge magnitudes then give $m_Y/m_X=2$ and equal speeds give $K_Y/K_X=2$ The magnetic force is perpendicular to the motion, so neither particle’s kinetic energy changes during its semicircle.

A: Gets the charge-to-mass ratio right but assumes equal speeds mean equal kinetic energies, overlooking the different masses.
C: Treats charge-to-mass ratio as directly proportional to radius; this also gives the inverse mass and kinetic-energy ratio.
D: Reverses the charge-to-mass ratio’s dependence on radius but correctly uses $K\propto m$ at equal speed.
Remember: a magnetic force changes a particle’s direction, not its kinetic energy.

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