Neco 2024 Physics Question 29

Practice objective / multiple choice question 29 from the 2024 Neco Physics examination.

Neco 2024 Physics Objective / Multiple Choice Hard Difficulty
Question 29 NECO • 2024 • PHYSICS • objective

A rectangular conducting loop of wire has dimensions \( 0.10\) m by \( 0.20\) m and a resistance of \( 2.0\) Ω. It lies in a uniform magnetic field of magnitude \( 0.50\) T directed into the page. The loop is pulled out of the field at a constant speed \( v \) so that the side of length \( 0.20\) m exits the field first.\n\n(a) Express the induced emf \( \varepsilon \) in terms of \( v \).\n\n(b) If the loop is made of copper with resistivity \( \rho = 1.68\times10^{-8}\ \Omega\,\text{m} \) and cross‑sectional area \( 1.0\times10^{-6}\ \text{m}^2 \), determine the speed \( v \) required for the average power dissipated in the loop while it is leaving the field to be \( 0.50\) W.\n\n(c) Using the speed found in part (b), calculate the induced emf, the time taken for the loop to leave the field completely, and the total thermal energy generated in the resistor.

Answer Options

A. v = 0.750 m·s⁻¹; ε = 0.0750 V; t = 0.150 s; thermal energy = 0.0750 J
B. v = 0.850 m·s⁻¹; ε = 0.0850 V; t = 0.170 s; thermal energy = 0.0850 J
Correct Answer C. v = 0.710 m·s⁻¹; ε = 0.0710 V; t = 0.141 s; thermal energy = 0.0705 J
D. v = 0.600 m·s⁻¹; ε = 0.0600 V; t = 0.120 s; thermal energy = 0.0600 J
Correct Answer
C
Correct Option:
v = 0.710 m·s⁻¹; ε = 0.0710 V; t = 0.141 s; thermal energy = 0.0705 J

About This Question

This is Neco 2024 Physics Question 29. It is one of the objective questions from the 2024 Neco Physics examination.

Difficulty level: Hard .

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