IB Physics SL D 3 5 Force on Current Conductor Questions

Calculate magnetic force on current-carrying conductors and coils, determine its direction, and apply it to motors, moving rods and electromagnetic systems.

Exam points

calculate the magnetic force on a current-carrying conductor using field strength, current, conductor length and angle
determine force direction on wires, coils and loops from current and magnetic-field directions, including torque or rotation effects
explain how charge carriers produce force on conductors and analyse motor, moving-rod, force-balance and force-per-unit-length evidence

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Study D.3.5—Force on current conductor conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

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The long straight conductor is formed into a coil consisting of two separate turns, X and Y . The coil hangs with its axis vertical.

Figure for Question IB Physics SL D 3 5 Force on Current Conductor Questions question 1 — IB Physics SL

Assume that the turns of the coil each behave as a long straight conductor.

The current in the coil is 15 A and the circumference of one turn is 0.48 m . In order to restore X and Y to their original separation, a mass of $2.8 \times 10^{-4} \mathrm{~kg}$ is suspended from turn Y . Estimate the magnetic field strength at X due to Y .

A3. This question is about radioactivity.

Caesium-137 ( ${ }_{55}^{137} \mathrm{Cs}$ ) is a radioactive waste product with a half-life of 30 years that is formed during the fission of uranium. Caesium-137 decays by the emission of a beta-minus $\left(\beta^{-}\right)$ particle to form a nuclide of barium (Ba).

$F=\left(0.280 \times 10^{-3} \times 9.81=\right) 2.75 \times 10^{-3} \mathrm{~N}$ ;
$B=\frac{F}{I l}$ or correct substitution $2.75 \times 10^{-3}=B \times(15) \times 0.48$ ;
$B=\left(\frac{2.75 \times 10^{-3}}{15 \times 0.48}=\right) 3.8 \times 10^{-4} \mathrm{~T} ;$

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IB Physics SL D 3 5 Force on Current Conductor Questions question 1