An electron travels into a 0.5 T magnetic field perpendicular to its path, where it moves in a circular arc of diameter 0.028 m. What is the kinetic energy of the electron?

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sampsonola sampsonola · Answer 1 · 2020-02-28T07:04:18+01:00

Answer:

6.89 × 10⁻¹³ Joules

Explanation:

Radius of the circular arc = diameter / 2 = 0.028 m / 2 = 0.014 m

radius of the circular arc = mev /qB

me = 9.11 × 10⁻³¹ kg ( mass of an electron)

q of an electron = 1.6 × 10⁻¹⁹ C

B = 0.5 T

v velocity of the electron = r qB / me = (0.014 m × 1.6 × 10⁻¹⁹ C × 0.5 T) / (9.11 × 10⁻³¹ kg ) = 0.00123 × 10 ¹² m/s = 1.23 × 10⁹ m/s

Kinetic energy = 1/2 mv² = 0.5 × 9.11 × 10⁻³¹ kg × (1.23 × 10⁹ m/s)² = 6.89 × 10⁻¹³ Joules

okpalawalter8 okpalawalter8 · Answer 2 · 2022-03-29T15:42:14+02:00

The kinetic energy of the electron is mathematically given as

K.E= 6.89* 10^{-13} Joules

Question Parameter(s):

An electron travels into a 0.5 T magnetic field

circular arc of diameter 0.028 m.

Generally, the equation for the radius of the circular arc is mathematically given as

R=d/2

R 0.028 m / 2

R= 0.014 m

Where

R= mev /qB

Hence

v = R qB / me

V=(0.014 m * 1.6 × 10^{19} C × 0.5 T)/(9.11 × 10{-31} kg )

V= 0.00123 * 10 ^{12} m/s

V= 1.23 × 10^9 m/s

In conclusion, The Kinetic energy

K.E= 0.5 mv^2

K,E= 0.5 * 9.11 * 10^{31} kg * (1.23 * 10^9 m/s)^2

K.E= 6.89* 10^{-13} Joules