Question

An electron moves in a region where the magnetic field is uniform and has a magnitude of 80 μT. The electron follows a helical path which has a pitch of 9.0 mm and a radius of 2.0 mm. What is the speed of this electron as it moves in this region?

Answer 1

Answer:

3.4 x 10⁴ m/s

Explanation:

Consider the circular motion of the electron

B = magnetic field = 80 x 10⁻⁶ T

m = mass of electron = 9.1 x 10⁻³¹ kg

v  = radial speed

r = radius of circular path = 2 mm = 0.002 m

q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

For the circular motion of electron

qBr = mv

(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v

v = 2.8 x 10⁴ m/s

Consider the linear motion of the electron :

v' = linear speed

x = horizontal distance traveled = 9 mm = 0.009 m

t = time taken = $\frac{2\pi m}{qB}$ = $\frac{2\pi (9.1\times 10^{-31})}{(1.6\times 10^{^{-19}})(80\times 10^{-6})}$ = 4.5 x 10⁻⁷ sec

using the equation

x = v' t

0.009 = v' (4.5 x 10⁻⁷)

v' = 20000 m/s

v' = 2 x 10⁴ m/s

Speed is given as

V = sqrt(v² + v'²)

V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)

v = 3.4 x 10⁴ m/s