AP EAMCET · PHYSICS · Magnetic Effects of Current
A charge \(q\) moving in a circle of radius \(r\) metre makes \(n\) rev/s. Magnetic field at the centre of the circle is
- A \(\frac{2 \pi q}{n r} \times 10^{-7} \mathrm{NA}^{-1} \mathrm{~m}^{-1}\)
- B \(\frac{2 \pi q}{r} \times 10^{-7} \mathrm{NA}^{-1} \mathrm{~m}^{-1}\)
- C \(\frac{2 \pi n q}{r} \times 10^{-7} \mathrm{NA}^{-1} \mathrm{~m}^{-1}\)
- D \(\frac{2 \pi \mathrm{q}}{r} \mathrm{NA}^{-1} \mathrm{~m}^{-1}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 \pi n q}{r} \times 10^{-7} \mathrm{NA}^{-1} \mathrm{~m}^{-1}\)
Step-by-step Solution
Detailed explanation
Current associated due to the movement of charge \(q\) in circular path of radius \(r\) is given as \(\begin{aligned} I & =\frac{q}{T}=\frac{q}{\left(\frac{1}{n}\right)} \quad\left[\because T=\frac{1}{n}\right] \\ & =q n \end{aligned}\) \(\therefore\) Magnetic field at the…
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