TS EAMCET · Physics · Magnetic Effects of Current
An electron beam travels with a velocity of \(1.6 \times 10^7 \mathrm{~ms}^{-1}\) perpendicularly to magnetic field of intensity \(0.1 \mathrm{~T}\). The radius of the path of the electron beam \(\left(m_e=9 \times 10^{-31} \mathrm{~kg}\right.\) )
- A \(9 \times 10^{-5} \mathrm{~m}\)
- B \(9 \times 10^{-2} \mathrm{~m}\)
- C \(9 \times 10^{-4} \mathrm{~m}\)
- D \(9 \times 10^{-3} \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(9 \times 10^{-4} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
In a perpendicular magnetic field, the radius of circular path travelled by electron beam is…
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