Question

An electron travels at 2.0 x 107 m/s in a plane perpendicular to a 0.1 T magnetic field. What is the radius and period of the electron? (hint: recall centripetal force and tangential velocity)

Answer 1

Answer:

r = 0.001137 m = 1.137 mm

T = 3.57 x 10⁻¹⁰ s

Explanation:

In order for the electron to remain in a fixed circle centripetal force must be equal to the magnetic force:

$Centripetal\ Force = Magnetic\ Force\\\frac{mv^2}{r} = qvB\ Sin\theta\\\\r = \frac{mv^2}{qvB\ Sin\theta} = \frac{mv}{qB\ Sin\theta}$

where,

r = radius = ?

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of electron = 2 x 10⁷ m/s

q = charge on electron= 1.6 x 10⁻¹⁹ C

B = Magnetic Fild Strength = 0.1 T

θ = Angle between velocity and magnetic field = 90° (perpendicular)

$r = \frac{(9.1\ x\ 10^{-31}\ kg)(2\ x\ 10^7\ m/s)}{(1.6\ x\ 10^{-19}\ C)(0.1\ T)Sin90^o}$

r = 0.001137 m = 1.137 mm

Now, for the period of the electron:

$v = \frac{2\pi r}{T}\\\\T = \frac{2\pi r}{v}$

where,

T = Time period required o cover a distance equal to cirumference = ?

$T = \frac{2\pi(0.001137\ m)}{2\ x\ 10^7\ m/s}$

T = 3.57 x 10⁻¹⁰ s