Question

An electron enters a magnetic field of 0.43 T with a velocity perpendicular to the direction of the field. At what frequency does the electron traverse a circular path? ( m el = 9.11 × 10-31 kg, e = 1.60 × 10-19 C)

Answer 1

Answer:

Frequency, $f=1.2\times 10^{10}\ Hz$

Explanation:

It is given that,

Magnetic field, B = 0.43 T

We need to find the frequency the electron traverse a circular path. It is also known as cyclotron frequency. It is given by :

$f=\dfrac{qB}{2\pi m}$

$f=\dfrac{1.6\times 10^{-19}\ C\times 0.43\ T}{2\pi \times 9.11\times 10^{-31}\ Kg}$

$f=1.2\times 10^{10}\ Hz$

Hence, this is the required solution.