AP EAMCET · PHYSICS · Magnetic Effects of Current
When a helium nucleus makes a full rotation of a circle of radius \(0.8 \mathrm{~m}\) in \(2.5 \mathrm{~s}\), then the value of magnetic field \(B\) at the centre of the circle will be
- A \(4 \pi \times 10^{-25} \mathrm{~T}\)
- B \(2 \pi \times 10^{-26} T\)
- C \(4 \pi \times 10^{-26} \mathrm{~T}\)
- D \(2 \pi \times 10^{-25} \mathrm{~T}\)
Answer & Solution
Correct Answer
(C) \(4 \pi \times 10^{-26} \mathrm{~T}\)
Step-by-step Solution
Detailed explanation
Charge on helium nucleus, \[ \begin{aligned} q & =2 e=2 \times 1.6 \times 10^{-19} \\ & =3.2 \times 10^{-19} \mathrm{C} \\ r & =0.8 \mathrm{~m}, \mathrm{~T}=2.5 \mathrm{~s} \end{aligned} \] Current associated due to one rotation of helium nucleus,…
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