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IB DP Physics Mock Exam HL Paper 1B Set 2 - 2025 Syllabus

IB DP Physics Mock Exam HL Paper 1B Set 2

Prepare for the IB DP Physics Mock Exam HL Paper 1B Set 2 with our comprehensive mock exam set 2. Test your knowledge and understanding of key concepts with challenging questions covering all essential topics. Identify areas for improvement and boost your confidence for the real exam

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Question 

A group of students aims to determine the horizontal component \(B_{H}\) of the Earth’s magnetic field. They position a magnetic needle halfway between two coils. When no current flows, the needle aligns in the North–South direction. When a current is switched on, the magnetic field generated by the coils causes the needle to deflect.
The students must choose a horizontal separation between the coils (\(2r\), \(r\), or \(0.5r\)). The graphs below illustrate how the magnetic field strength varies with distance for each separation.
(a) State and explain which coil separation should be selected for this investigation.
(b) Explain why the students align the axis XX′ of the coils in the east–west direction.
(c) The magnetic needle settles at an angle of \(24^{\circ}\) to XX′. Using the graphs and the diagram below, determine the value of \(B_{H}\).
▶️ Answer/Explanation
Detailed solution

(a)
The students should choose the separation \(r\). This separation produces the most uniform magnetic field between the coils, as indicated by the flat region of the field curve, ensuring a nearly constant field across the magnet.

(b)
The axis XX′ is aligned east–west so that the magnetic field produced by the coils (\(B_{\text{coils}}\)) is perpendicular to the Earth’s horizontal magnetic field (\(B_{H}\)), which lies in the north–south direction. This perpendicular arrangement allows the fields to be combined vectorially to determine \(B_{H}\).

(c)
From the vector relationship, with \(B_{\text{coils}}\) along XX′ and \(B_{H}\) perpendicular to it:
\[\tan(24^{\circ}) = \frac{B_{H}}{B_{\text{coils}}}\]
From the graph corresponding to separation \(r\), the magnetic field at the centre is approximately \(7.5 \times 10^{-5}\,\text{T}\).
\[B_{H} = (7.5 \times 10^{-5}) \times \tan(24^{\circ}) \approx 3.3 \times 10^{-5}\,\text{T}\]

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