Answer:
the angle between the electron's velocity and the magnetic field is 19⁰
Explanation:
Given;
magnitude of the magnetic field, B = 83 x 10⁻⁵ T
acceleration of the electron, a = 34 x 10¹³ m/s²
speed of the electron, v = 72 x 10⁵ m/s
mass of electron, m = 9.11 x 10⁻³¹ kg
The magnetic force experienced by the electron is calculated as;
F = ma = qvB sinθ
where;
q is charge of electron = 1.602 x 10⁻¹⁹ C
θ is the angle between the electron's velocity and the magnetic field.
[tex]sin(\theta ) = \frac{ma}{qvB} \\\\sin(\theta ) = \frac{(9.11\times 10^{-31})(34\times 10^{13})}{(1.602\times 10^{-19})\times (72\times 10^5) \times (83 \times 10^{-5})} \\\\sin(\theta ) = 0.3235\\\\\theta =sin^{-1}(0.3235)\\\\\theta =18.9^0[/tex]
[tex]\theta \approx 19^ 0[/tex]
Therefore, the angle between the electron's velocity and the magnetic field is 19⁰
