Electron in Magnetic Field
Question: An electron (mass \( m \), kinetic energy \( E \)) enters a uniform magnetic field \( B \) perpendicularly. Its frequency is:
(a) \( \frac{eE}{qvB} \)
(b) \( \frac{eB}{2\pi m} \)
(c) \( \frac{2\pi m}{eB} \)
(d) \( \frac{2m}{eBE} \)
Solution
Step 1: Magnetic force provides centripetal force
\( qvB = \frac{mv^2}{r} \) ⇒ \( r = \frac{mv}{qB} \)
Step 2: Relationship between velocity and frequency
\( v = \omega r = 2\pi f r \)
Step 3: Substitute \( r \) from Step 1 into Step 2
\( v = 2\pi f \left(\frac{mv}{qB}\right) \) ⇒ \( f = \frac{qB}{2\pi m} \)
Step 4: For electron (\( q = e \))
\( f = \frac{eB}{2\pi m} \)
Answer: Option (b) is correct
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