Question

A cosmic ray electron moves at 7.50×106 m/s perpendicular to the Earth’s magnetic field at an altitude where field strength is 1.00×10−5 T . What is the radius of the circular path the electron follows?

Answer 1

Answer:

4.266 m.

Explanation:

Given,

Speed of the electron = 7.50 x 10⁶ m/s

Magnetic field, B = 1 x 10⁻⁵ T

radius of circular path = ?

Using formula of radius of curvature of the charge particle

$r = \dfrac{mv}{qB}$

m is mass of electron

q is charge of electron

$r = \dfrac{9.11 \times 10^{-31}\times 7.50\times 10^6}{1.6\times 10^{-19}\times 1 \times 10^{-5}}$

r = 4.266 m

Hence, radius of circular path is equal to 4.266 m.