NESA Physics The Motor Effect

1 sample question with marking guides and sample answers · Avg. score: 49.2%

Q10
2024
QCAA
Paper 1
1 mark
Q10
1 mark

An experiment was conducted to determine the force experienced by an 85 cm wire with a 2.4 A current flowing through it in an external magnetic field. It was rotated through varying angles within the magnetic field such that data analysis identified the relationship F=0.0306sinθF = 0.0306 \sin \theta.

What is the order of magnitude of the strength of the external magnetic field?

A

10410^{-4} T

B

10210^{-2} T

C

10210^{2} T

D

10410^{4} T

Reveal Answer
A

10410^{-4} T

This value is too small. The calculated magnetic field strength is approximately 0.015 T0.015 \text{ T}, which is two orders of magnitude larger than 104 T10^{-4} \text{ T}.

B

10210^{-2} T

Correct Answer

Comparing the experimental relationship F=0.0306sinθF = 0.0306 \sin \theta to the theoretical formula F=ILBsinθF = ILB \sin \theta, we see that ILB=0.0306ILB = 0.0306. Solving for BB gives B=0.03062.4×0.850.015 TB = \frac{0.0306}{2.4 \times 0.85} \approx 0.015 \text{ T}, which is on the order of 102 T10^{-2} \text{ T}.

C

10210^{2} T

This value is significantly larger than the actual magnetic field. The calculation yields B0.015 TB \approx 0.015 \text{ T}, whereas 102 T10^2 \text{ T} represents an extremely strong magnetic field not supported by the data.

D

10410^{4} T

This value is far too large. The calculated field strength is approximately 1.5×102 T1.5 \times 10^{-2} \text{ T}, which is six orders of magnitude smaller than 104 T10^4 \text{ T}.

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