IB Physics HL D.3 Motion in electromagnetic fields questions

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Question 1

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.

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