AP EAMCET · PHYSICS · Magnetic Effects of Current
The radius of the path of an electron moving at a speed of \(3.2 \times 10^7 \mathrm{~ms}^{-1}\) in a magnetic field of \(6 \times 10^{-4} \mathrm{~T}\) perpendicular to it is (mass of electron is \(9 \times 10^{-31} \mathrm{~kg}\) and charge of electron is \(\left.1.6 \times 10^{-19} \mathrm{C}\right)\)
- A \(22.4 \mathrm{~cm}\)
- B \(13 \mathrm{~cm}\)
- C \(30 \mathrm{~cm}\)
- D \(39 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(C) \(30 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{9.1 \times 10^{-31} \times 3.2 \times 10^7}{1.6 \times 10^{-19} \times 6 \times 10^{-4}}=30 \mathrm{~cm}\)
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