Question

A proton moves in a circular path of the same radius as a cosmic ray electron moving at 6.5 × 10^6 m/s perpendicular to the Earth’s magnetic field at an altitude where the field strength is 1.0 × 10^(-5) T.
What is the radius of the circular path the electron follows?

Answer 1

Answer:

3.7 m

Explanation:

Velocity, v = 6.5 x 10^6 m/s

magnetic field, B = 1 x 10^-5 T

charge, q = 1.6 x 10^-19 C

mass, m = 9.1 x 10^-31 kg

Let r be the radius of the path.

$r=\frac{mv}{Bq}$

$r=\frac{9.1\times 10^{-31}\times 6.5 \times 10^{6}}{1\times 10^{-6}\times 1.6\times 10^{-19}}$

r = 3.7 m

Answer 2

Answer:

3.7 m

Explanation:

We are given that

Mass of electron,m=$9.1\times 10^{-31} kg$

$Speed of electron,v=6.5\time 10^6 m/s$

Magnetic filed,B=$1.0\times 10^{-5} T$

Charge on electron,q=$1.6\times 10^{-19} C$

We have to find the radius of circular path the electron follows.

We know that

Radius of circular path,r=$\frac{mv}{qB}$

Using the formula

$r=\frac{9.1\times 10^{-31}\times 6.5\times 10^6}{1.6\times 10^{-19}\times 1\times 10^{-5}}$

$r=3.7 m$

Hence, the radius of circular path followed by electron=3.7 m