COMEDK · Physics · 21. Magnetic Effects of Current
A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5 A, the cross-sectional area of the wire is \(10^{-5} \text{ m}^2\), and the free electron density is \(10^{29} \text{ m}^{-3}\), then the average force on each electron in the coil due to the magnetic field is:
- A \(2.5 \times 10^{-25} \mathrm{~N}\)
- B \(4.5 \times 10^{-25} \mathrm{~N}\)
- C \(5 \times 10^{-25} \mathrm{~N}\)
- D \(5.5 \times 10^{-25} \mathrm{~N}\)
Answer & Solution
Correct Answer
(C) \(5 \times 10^{-25} \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
Force on one electron: \(F = ev_d B\) From \(I = nAev_d \Rightarrow ev_d = \dfrac{I}{nA}\) \(F = \dfrac{IB}{nA} = \dfrac{5 \times 0.10}{10^{29} \times 10^{-5}} = \dfrac{0.5}{10^{24}} = 5 \times 10^{-25}\,\text{N}\) (Number of turns and radius do not affect the force on an…
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