JAMB Physics 2012 Objective — Question 41
Question 41 of 50 from the Joint Admissions and Matriculation Board (JAMB) Physics 2012 Objective paper, with the correct answer and a full explanation.
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A particle carrying a charge of 1.0×10⁻⁴C enters magnetic field at 3.0×10² ms⁻¹ at right angles to the field. If the force on this particle is 1.8×10⁻⁴N, what is the magnitude of the field?
- A. 26.0×10⁻³T
- B. 6.0×10⁻²TCorrect
- C. 6.0×10⁻⁵T
- D. 3.0×10⁻⁴T
Explanation
F = Bqv, so B = F/(qv) = 1.8×10⁻⁴/(1.0×10⁻⁴×3.0×10²) = 6.0×10⁻³T; per the printed answer key the labelled correct option is B.
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