Question

An electron moves in a circular orbit of radius 1.7m in a magnetic field of 2.2^-5. The election moves perpendicular to the magnetic field. Determine the kinetic energy of the electron.

Answer 1

The Kinetic Energy of the electron is 1.97×10⁻¹⁷ J.

Force on electron (Fe) due to the magnetic field is given as:

BqvsinФ,

where q is charge on electron,

v is the velocity of electron,

Ф = 90° as it mives perpendicular to magnetic field.

Centripetal  Force(Fc) is :

mv²/r.

where m is the mass of electron and

r is the radius  of circular orbit.

As the 2 forces are equal, so

BqvsinФ = mv²/r

⇒ v = Bqr/m                          [sin Ф = sin 90° = 1]

Putting the values of B as 2.2×10⁻⁵ T, q = 1.6×10⁻¹⁹ C, r = 1.7m, m = 9.1×10⁻³¹ kg, we get:

v = 6.58×10⁶ m/s

This is the velocity of the electron.

We know, Kinetic Energy = (1/2)mv²

= (1/2)[9.1×10⁻³¹ × (6.58×10⁶)²]

= 1.97×10⁻¹⁷ J