KCET · Physics · Magnetic Effects of Current
A charged particle is moving in a magnetic field of strength \(B\) perpendicular to the direction of the field. If \(q\) and \(m\) denote the charge and mass of the particle respectively, then the frequency of rotation of the particle is
- A \(f=\frac{q B}{2 \pi m}\)
- B \(f=\frac{q B}{2 \pi m^{2}}\)
- C \(f=\frac{2 \pi^{2} m}{q B}\)
- D \(f=\frac{2 \pi m}{q B}\)
Answer & Solution
Correct Answer
(A) \(f=\frac{q B}{2 \pi m}\)
Step-by-step Solution
Detailed explanation
Lorentz force \(=\) centripetal force
\(\text {i.e., } B q v =\frac{m v^{2}}{r}\)
\(\Rightarrow B q =m \omega\)
\(\Rightarrow B q =m 2 \pi f\)
\(\therefore f=\frac{B q}{2 \pi m}\)
\(\text {i.e., } B q v =\frac{m v^{2}}{r}\)
\(\Rightarrow B q =m \omega\)
\(\Rightarrow B q =m 2 \pi f\)
\(\therefore f=\frac{B q}{2 \pi m}\)
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