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प्रश्न
A tightly-wound, long solenoid carries a current of 2.00 A. An electron is found to execute a uniform circular motion inside the solenoid with a frequency of 1.00 × 108 rev s−1. Find the number of turns per metre in the solenoid.
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उत्तर
Given:
Magnitude of current in the solenoid, i = 2 A
Frequency of the electron, \[f = 1 \times {10}^8\] rev/s
Mass of the electron, \[\text{ m }= 9 . 1 \times {10}^{- 31}\] kg
Charge of the electron, \[q = 1 . 6 \times {10}^{- 19} C\]
We know that the magnetic field inside a solenoid is given by
B = µ0ni
If a particle executes uniform circular motion inside a magnetic field, the frequency of the particle is given by
\[f = \frac{qB}{2\pi m}\]
\[ \Rightarrow B = \frac{2\pi mf}{q}\]
\[ \Rightarrow \mu_0 ni = \frac{2\pi mf}{q} [\text{ Using } (1)]\]
\[ \Rightarrow n = \frac{2\pi mf}{\mu_0 qi}\]
\[ = \frac{2\pi \times 9 . 1 \times {10}^{- 31} \times 1 \times {10}^8}{4\pi \times {10}^{- 7} \times 1 . 6 \times {10}^{- 19} \times 2}\]
\[ = 1420 \] turns/m
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